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IN  MEMOR1AM 
FLOR1AN  CAJOR1 


IMPROVED  EDITION  WITH  QUESTIONS. 

A 

SHORT  SYSTEM  OF 

PRACTICAL,  ARITHMETIC, 

COMPILED 

FROM  THE  BEST  AUTHORITIES,; 

f      •  TO    WHICH   IS   ANNEXED 

A  SHORT  PLAN  OF  BOOK-KEEPING. 


*  THE    WHOLE   DESIGNED 

¥  •**> 

.'  i  FOR  TEE  USE  OF  SCHOOLS.  ~& 


BY  WILLIAM  KINNE,  A.  M.  !  ' 


asirftfon, 


WITH  QUESTIONS 

ON    EVERY    PART    OF    ARITHMETIC,    AND    A    COMPENDIOUS 
SYSTEM    OF    TAX    MAKING. 


REVISED,  CORRECTED,  AND  GREATLY  ENLARGED, 

BY  DANIEL  ROBINSON. 


HA  LL  O  WELL  : 

PRINTED  AWD  PUBLISHED  BY  GLAZIER,  MASTERS  &  Co. 

Sold  by  them  at  the  Hailowell  Bookstore,  No.  I,  Kennebec-Row  ,  and  by  all 

the  Booksellers  in  the  State. 


1831. 


DISTRICT  OF  MAINE,  si. 

BE  IT  REMEMBERED,  That  on  the  seventh  day  of  December,  in  the 
year  of  our  Lord  one  thousand  eight  hundred  and  twenty-two,  and  the  forty  - 
seventh  year  of  the  Independence  of  the- United  States  of  America,  GOOD- 
ALE,  GLAZIER  &  COMPANY  of  the  District  of  Maine,  have  deposited  in  this 
office,  the  title  of  a  Book,  the  right  whereof  they  claim  as  proprietors — in  the 
words  following,  viz. : — "  Improved  edition  with  questions.  A  short  system 
'  of  Practical  Arithmetic,  compiled  from  the  best  authorities  ;  to  which  is 
1  annexed  a  short  plan  of  Book-keeping.  The  whole  designed  for  the  use  of 
'  Schools.  By  William  Kinne,  A,  M.  Fourth  edition,  with  qaestions  on  every 
*  part  of  Arithmetic,  and  a  compendious  system  of  Tax  making.  Revised, 
'  corrected  and  greatly  enlarged,  by  Daniel  Robinson.  Hallowell,  printed 
'  and  published  'by  Goodale,  Glazier  &  Co."— In  conformity  to  the  Act  of 
the  Congress  of  the  United-States,  entitled,  "  An  Act  for  the  encouragement 
of  learning,  by  securing  the  copies  of  maps,  charts,  and  books,  to  the  authors 
aad  proprietors  of  such  copies,  during  the  times  therein  mentioned  j  and  also 
to  an  Act,  entitled  "  An  Act  supplementary  to  an  Act,  entitled  an  Act  for  the 
encouragement  of  learning,  by  securing  the  copies  of  maps,  charts  and  books 
to  the  authors  and  proprietors  of  such  copies,  during  the  times  therein  men- 
tioned, and  extending  the  benefits  thereof  to  the  arts  of  designing,  engraving, 
arid  etching  historical  and  other  prints." 

JOHN  MUSSEY,  JR.,  Clerk  of  the  District  Court  of  Maine. 

A  true  copv  as  of  record. 

Attest,  J.  MUSSEY,  JR.,  Clerk  D.  C.  Maine. 


Advertisement  to  the  Seventh  Edition. 

J  has  been  the  primary  purpose,  in  each  improved  Edition  of  this  Work, 
to  render  it  more  and  more  plain  and  practical,  while  it  should  embrace  every 
useful  rule  and  question  which  might  occur  in  the  ordinary  business  transac- 
tions of  life.  To  effect  this  object,  neither  time  nor  thought  has  been,  in  any 
wise,  niggardly  expended.  Whatever  was  judged  to  be  wanted,  to  charac- 
terize it  as  a  plain,  practical,  and  useful  system,  has  been  amply,  though 
gradually,  supplied.  In  the  edition  now  presented  to  the  public,  part  of  the 
questions  in  which  avoirdupois  weight  is  concerned,  has  been  writtenjanew,  or 
so  altered  as  to  allow  25  pounds  only  to  the  quarter  of  a  hundred  weight  ; 
because  this  practice  now  generally  obtains  in  business,  among  merchants 
and  traders  in  the  United-States,  and  has  moreover  been  established  in  Maine 
by  legislative  enactment.  Considerable  new  matter  also  has  been  crowd- 
ed into  the  volume,  and  a  small  portion  of  the  old  withdrawn.  Errour  has 
been  diligently  sought  for  and  corrected;  and,  it  is  confidently  believed,  is 
now  nowhere  to  be  found  on  its  pages.  Considered  as  an  Epitome,  whether 
it  be  susceptible  of  any  farther  degree  of  improvement,  may  be  reasonably  v 
^  questioned.  The  hope  is,  therefore,  indulged,  that,  though  the  tongue  of  the  j)/ 

/7/Ycaptious  caviller  should  blazon  defects  for  which  others  might  search  in  vain  ; 

C*»i//  vet  the  eye  of  the  candid  critic  will  see  nothing  in  this  compendium  which 
reason  and  truth  would  long  hesitate  to  .approve. 
Gardiner,  August  1,  1828. 


To  adapt  tliis  work  to  the  easy  use  of  Instructors,  I  have  endeavoured  to 
simplify  the  definitions  and  rules,  so  as  to  render  them  as  familiar  and  concise 
as  the  nature  of  the  subject  admits.  At  the  san-ie  time,  I  have  very  consider- 
ably enlarged  the  Original  System,  by  the  insertion  of  a  far  greater  number 
of  practical  examples,  especially  in  the  grourid-ruies,  and  by  the  introduction 
of  many  new  rules,  in  order  to  furnish  our  Schools  with  a  methodical  and 
comprehensive  Treatise  of  Practical  Arithmetic. 

Works  of  this  kind  have  too  often  abounded  with  abstruse  and  intricate 
questions,  more  puzzling  than  beneficial  to  the  learner. — And  some  authors 
have  dwelt  too  much  on  those  of  a  trifling1  nature,  which,,  when  understood, 
afford  no  useful  knowledge.  To  avoid  these  extremes,  to  feed  and  invigorate 
the  mind,  and  thus  form  onr  youth  for  entering,  with  fair  promise,  on  the  pur- 
suits of  active  life,  have  been  my  principal  aims;  in  preparing  this  edition  for 
the  press. 

Most  of  the  former  demonstrations  have  been  omitted,  as  being  little  suited 
to  enlighten  the  pupil,  and  as  excluding,  in  such  compends,  matter  much  more 
conducive  to  the  purpose  of  his  instruction.  The  book-keeping,  also,  has 
been  somewhat  abridged,  for  the  admissioi\of  other  matter  j  yet  enough,  it  is 
conceived,  has  been  retained  to  give  the  student  no  very  imperfect  idea  of  this 
branch  of  learning.  Besides  what  has  been  substituted  in  place  of  this  excluded 
matter,  no  fewer  than  51  pages  have  been  added  to  the  last  edition.  To  the 
whole  have  been  prefixed  brief  questions  on  all  the  most  important  parts  of 
Arithmetic.  But,  instead  of  entering  into  a  detail  of  these  enlargements,  I 
beg  leave  to  refer  the  reader  to  the  table  of  contents,  or  to  the  pages  of  the 
work  itself. 

During  the  many  }'ears  that  I  have  devoted  to  the   instruction  of  youth  in 
Arithmetic,  1  have  used  various  systems,  all  of  which  havejust  claims  to  sci- 
entific merit.     The  authors,  however,  have,  generally,  appeared  to  be  defi- 
cient in  an  important  point — the  practical  teacher's  experience.      They  have       • , 
(  Jlgg^^lJiClLtQQ- spa  ring  of  examples^more^  esneciajjy  in   the  A^^Biles^  __The  JLL&Ll 
[    coisequeSce"!^  tfiaTlnVsch^Dlaris^lmrried  ihroug ;  h i  me^eTundarnenta  1  rules  fas-  \J  •!*' ^-- 
ter  than  his  comprehension  and  proficiency  would  justify.    To  obviate  this  ob- 
jection, has  been  another  design  in  the  present  undertaking. 

Considering  that,  to  attain  a  thorough  knowledge  of  vulgar  fractions,  is 
usually  too  difficult  a  task  for  young  students,  whose  progress  in  Arithmetic 
has  extended  only  to  compound  division,  and  that  the  difficulty  frequently  re- 
sults in  their  utter  discouragement  j  I  have,  therefore,  deemed  it  most  advise- 
abie  and  advantageous  to  transfer  these  fractions  (except  two  or  three  prob- 
lems .jntrod^tor^to_dgc»nals)  beyond  equation  o*  payments.^  But,  asjfocimat 
fracTibns'may  be  more  easily  acquired,  are  more  simple,  useful,  ana  necessa-' 
ry,  and  ara  sooner  wanted  in  the  practical  branches  of  numbers,  1  have 
thought  it  expedient  to  let  them  occupy  that  part  of  the  work  which  they  did 
in  former  editions.  The  other  rules  I  have  likewise  aimed  so  to  arrange,  as 
to  give  precedence  to  those  which  are  most  simple  and  necessary,  introducing 
the  more  abstruse  and  difficult  parts  last.  The  teacher,  however,  will  not 
consider  himself  as  being  obliged  to  adhere  strictly  to  this  arrangement.  He 
can,  notwithstanding,  take  the  rules  in  such  order  as  he  may  conceive  to  be 
the  most  proper.  D.  R. 


TABLE  OF  CONTENTS. 


Notation  or  Numeration, 
r  Addition, 
Subtraction, 

Multiplication, 
Division, 
Practical  Questions, 


Simple 


Tables 


Of 


and  Measures, 
Reduction, 

Application  of  Reduction, 
Federal  Monoy, 
Reduction  of  Federal  Money, 
Practical  Questions  in  ) 
Federal  Money,        J 
Addition, 


Division, 
Average  Judgement, 
Duodecimals, 
.  Three  Problems  in  Vulgar 

Fractions, 
Decimal  Fractions, 
Reduction  of  Currencies, 
Rule  of  Three, 
Practice, 
Tare  and  Tret, 
Double  Rule  of  Three, 
Conjoined  Proportion, 
Barter, 
Loss  and  Gain, 


Simple  Interest, 
Short  Practical  Rules, 
To  compute  Interest  on  ) 

Notes,  Bonds,  &c. 
Compound  Interest, 
Commission, 
Insurance, 
Discount, 

Annuities  at  Simple  Interest, 
Equation  of  Payments, 
Exchange, 


PAGE\ 

PAGE. 

9 

12 

Vulgar  Fractions, 
Rule  of  Three  in  Vul.  Frac. 

147 
160 

15 

Rule  of  Three  in  Dec.  Frac. 

161 

16 

Double  Rule  of  Three  in  ) 

_» 

21 

Vulgar  Fractions,       $ 

loZ 

25 

Simple  Interest  in  Decimals, 

•    163 

?     07 

Table  of  Ratios, 

163 

) 

Compound  Interest  in  Decima 

a,    167 

32 

Table  of  the  amount  of   £l, 

) 

37 

or  $1,  at  5  and  6  per  cent. 

/ 

38 

per  annum,  compound  inter- 

\ 

44 

est,  for  20  3Tears, 

; 

48 

Annuities  at  compound  interest 
Involution, 

,     168 
169 

48 

Evolution,  or  Extraction  ) 

17fl 

56 

of  Roots,               5 

I  /  U 

61 

66 

Application  and  Use  of  the  ) 
Square  Root.             \ 

173 

72 

The  Cube  Root, 

177 

73 

Application  and  Use  of  the  ) 

f. 

Cube  Root,                5 

7o 

Roots  of  Powers  in  general, 

179 

79 
88 

Progression     ^  Geolnet^ical1' 

181 

183 

98 

C    TVTftrJJol 

187 

J05 

Alligation    ,    }  Alternate 

188 

108 

p    .  .        (  Single, 

190 

111 

Position    <  i)ouble, 

191 

114 

Permutation  and  Combination, 

193 

115 

Miscellaneous  Questions, 

196 

117 

Measurement  of  Grindstones, 

201 

119 

122 

Mensuration    <  °^  s°Fds  ^^ 

202 
211 

123 

Cask  Gauging, 

219 

127 

Ships'  Tonnage  and  Length  > 
of  Masts,                      J 

221 

130 

Weight  of  Anchor  a  Cable  \ 

135 

may  sustain  —  Burthen  of  f 

999 

137 

Ships  from  their  Propor-  t 

££% 

137 

tions, 

138 

Wood  and  Bark  Measure, 

223 

140 

Assessing  Taxes, 

223 

141 

142 

Book-keeping1, 
Tables  of  value  of  gold, 

226 
236 

Topics  for  Examination  in  the  Arithmetic. 


What  is  Arithmetic?  What  are  the  four  fundamental  Rules  for  its  opera- 
tion ?  To  understand  these,  what  is  previously  necessary  ?  \Vhat  does  No- 
tation teach  1  How  many  characters  or  figures  are  employed  in  it  ?  By 
what  common  term  are  the  first  nine  called  ?  How  named  ?  What  does  the 
tenth  figure  denote  ?  Have  not  these  digits  a  local,  as  well  as  a  simple  value  1 
On  what  principle,  does  their  local  value  depend  ?  .  Denominate  the  names  of 
the  places,  according  to  their  order.  How  is  the  cipher  used,  in  connection 
with  the  significant  figures  ?  In  what  manner  are  large  numbers  divided? 
Name  the  places  *  and  read  each  line  of  numbers,  in  the  Numeration  Table. 
What  is  the  Rule  for  expressing  numbers  m  figures  when  above  nine  ?  Give 
the  Rule  to  'read  numbers.  What  is  Addition?  Simple?  Let  me  examine 
you  in  the  Table.  Recite  the  Rule  and  modes  of  Proof.  What  is  Subtrac- 
tion? When  Simple?  Name  its  numbers.  Let  me  examine  you  in  the  Ta- 
ble. Give  the  Rule  and  way  of  Proof.  What  is  Multiplication  ?  Name  its 
numbers  What  common  term  is  applied  to  \\\G  first  two  of  them  1  When  is 
it  Simple?  How  is  the  Table  used  ?  Let  me  examine  you  in  it.  When  is  it 
Cast  first?  Repeat  the  Rule.  When  Case  second?  Give  the  Rule,  and 
modes  of  Proof.  When  does  the  first  Case  of  Contractions  apply  1  Give  the 
Rule.  When  the  second  Case  ?  Tel)  the  Rule.  What  is  shown  by  Divi- 
sion 1  Name  its  numbers.  When  is  it  Simple  ?  Let  me  examine  you  in  the 
Table.  Repeat  the  Rule,  Notes,  and  modes  of  Proof.  Repeat  the  directions 
in  Case  first  of  Contractions.  What  Case  second  and  Rule  1  What  says 
Case  third  ?  What  its  Note  and  Rule  ?  Repeal  the  Money  Table.  That  of 
Troy  Weight.  Apothecaries'  Weight.  Avoirdupois  Weight.  Cloth  Meas- 
ure. Long;  Measure.  Square  Measure.  Cubic  Measure,  Dnj  Measure. 
Wine  Measure.  Ale  Measure.  Time.  Planetary  Motion.  .What  is  taught 
by  Reduction  1  Repeat  the  Rules  and  Proof.  Tell  me  the  mode  of  for- 
mation,' and  Tabled  Federal  Money.  Give  the  Ride  for  its  Addition  For 
its  Subtraction.  For  its  Multiplication.  Recite  ihe  J\'ote,  and  A  Short  Rn/e. 
Give  the  Rule  for  its  Division.  Tell  the  Short  Rule.  Give  the  Rules  for  the 
Reduction  of  Federal  Money.  What  is  the  DiiecLion  in  Case  first  for  chang- 
ing New-England  currency  to  Federal  Money  ?  In  case  second  1  In  Case 
third?  What  is  the  Ride  in  Case  first  for  "changing  Federal  Money  to 
New-England  currency  ?  In  Case  second  ?  Whk  is  taught  by  Compound 
Addition?  Give  the  Rule.  What  is  Compound)  fubf.raclf.on?  Toll  the 
Rule.  What  does  Compound  Multiplication  teach  ?  Give  the  Rule.  What 
is  the  direction  in  Case  second  1  What  in  Ca«e  third  1  What  does  Com- 
pound Division  teach  ?  Rehearse  the  Rule.  What  does  Case  second  direct  ? 
What  Case  tfu'rrf  ?  Tell  the  JVbte  before  examples  in  Average  Judgment, 
What  is  observed  of  Duodecimals  1  Give  the  Rule  for  multiplying  them. 
What  are  Fractions  ?  How  is  a  Vulgar  Fraction  represented?  Name  its 
parts.  What  is  shown  by  the  Denominator  ?  What  by  the  Numerator? 
When  is  a  fraction  in  its  lowest  terms?  Give  the  rule' for  Problem  first. 
What  is  the  intent  of  Problem  second  ?  Tell  its  rule.  What  of  Problem 
third  1  What  \isrule?  What  is  a  Decimal  Fraction?  How  expressed  ? 
What  determines  its  relative  value  ?  How  are  such  fractions  affected  by  ci~ 


TI  TOPICS    FOR    EXAMINATION. 

Reduction  ?  In  second  Case  ?  IB  third  1  In  fourth  1  In  fifth  ?  What  are 
the  rides  in  Cascjirst  of  Reduction  of  Currencies  ?  In  case'second  ?  third  ? 
fourth?  fifth?  What  are  the  rules  for  changing  Federal  Money  into  the 
Currencies  of  the  several  Slates  ?  What  for  changing  it  to  Canada  and  Nova 
Scotia  money  ?  What  to  that  of  Great  Britain  ?  What  does  the  Rule  of 
Three  teach  ?  Why  so  na?/^  ?  Why  called  Golden  Rule  ?  Give  the  rules, 
and  notes  for  its  operation.  What  is  Practice  ?  What  the  rule  in  Case  first  1 
What  says  Case  second,  and  what  its  rule  1  What  are  Tare  and  7Ve£  ?"  De- 
fine all  its  terms.  When  is  Casejirst  used,  and  what  its  rule  1  When  Case 
second,  and  what  the  rule  1  Case  third,  and  what  the  rule  ?  Fourth,  and 
what  the  rule  1  What  does  the  Double  Rule  of  Three  teach  ?  How  many 
terms  in  its  questions  ?  How  distinguished  ?  Give  the  rule  and  wtftes.  What 
is  Conjoined  Proportion  ?  When  is  Case  first  used,  and  what  its  rules  ? 
When  Case  second,  and  what  its  rule  1  What  is  Barter  ?  Its  rule?  What 
Loss  and  Gain?  In  what  instruct  Merchants  and  traders?  How  its  #z/es- 
/zorts  solved  ?  What  its  general  law  ?  What  is  Fellowship  1  Its  wse?  What 
Single  Fellowship  ?  Its  72z//e  ?  How  /?rweJ  ?  What  Double  Fellowship  ? 
Its  Rule  1  What  is  Interest  ?  What  the  fcg-^  interest  ?  Define  its  term. 
How  many  kinds  ?  What  Simple?  ]is  Rule  and  Note  ?  Tell  the  7\iW<?  of 
Aliquot  Farts  '?  What  the  Rule  for  Months,  at  6  per  cent.  ?  For  ctoys  at 
ditto  ?  What  the  Short  Practical  Rule  for  pounds,  &c.  at  ditto  ?  How  serve 
at  5  or  7  per  cent.  ?  What  the  Short  Rule  for  Federal  Moneij,  at  6  per  cent.  T 
How  proceed  at  5  or  7  per  cent.?  What  the^r-s^  Rule  to  compute  Interest  on 
Notes,  &c.  having  Endorsements  1  How  the  Rule  contracted  ?  What  the 
Rule  in  Massachusetts  ?  What  Compound  Interest  ?  What  the  Rule  1  What 
Commission  and  Brokerage?  What  Insurance?  What  Discount?  What 
Present  Worth  1  What  the  Rules?  What  the  /Jw/e  when  there  are  several 
sums  to  be  paid,  A:c.  ?  What  an  Annuity?  How  in  Arrears?  What 
meant  by  Amount  ?  What  by  Present  worth  1  How  is  the  Amount  found  at 
Simple  Interest  ?  How  the  JFVeserrf  Worth  1  What  Equation  of  payments  ? 
What  the  UM/C  ?  What  known  by  Exchange  ?  Tell  the  ToWe.  What  the 
Rides?  How  many  &iWs  of  Vulgar  Fractions?  What  a  Proper  one? 
Improper  ?  Single  ?  Compound  ?  Mixed  ?  How  turn  a  u'/io/e  number  to  a 
Fraction?  What  a  Complex  one?  What  does  a  fraction  denoted  What 
its  value  egwa/ to  ?  What  meant  by  Common  Measure  1  W7hat  by  Cornmon 
Multiple?  Give  ihe  Rules  of  Pvob\em first.  Of  Problem  second*  What  is 
Reduction  of  Vulgar  Fractions  ?  Repeat  the  /as*  part  of  the  rule  in  Case 
first.  What  the  Rule  in  Case  second  1  What  in  Case  third  ?  In  Case  fourth  ? 
Those  in  Case  fifth  1  Tell  the  rule  in  Case  sfri/i  ?  What  the  rw/e  in  Case 
eighth?  WTha~t  are  the  rule  and  m?fos  in  Addition  of  Vulgar  Fractions? 
What  in  Subtraction  ?  What  in  Multiplication  ?  What  in  Division  ?  How 
do  you  proceed  in  the  Rule  of  Three  in  Vulgar  Fractions*?  How  in  the 
Rule  of  Three  in  Decimals  1  How  in  the  Double  Rule  of  Three  hi  Vulgar 
Fractions?  Tel!  the  Table  of  Ratios  in  Simple  Interest  by  Decimals.  What 
is  Ratiol  How  do  you  find  the  Interest  1  What  the  Rule  in  Case  second  ? 
In  Case  third  ?  Case  fourth  ?  How  find  Interest  for  Days  ?  How  calculat- 
ed on  Cash  Accounts  where  partial  Payments  are  made  ?  How  find  Com-' 
pound  Interest  by  Decimals  ?  How  find  Amount  of  an  Annuity  at  Compound 
Interest  1  How  its  Present  Worth  at  ditto  ?  What  is  Involution  1  What  the 
/rrsf  Power  ?  What  the  second  ?  The  third  1  The  /o?*r/ft  ?  What  J£v0/u- 
"tt'ori  or  Extraction  of  Roots  ?  What  the  Root  1  Can  the  Root  of  any  num- 
ber be  found  ?  How  approximate  towards  it  ?  What  are  Roots  called  ? 
What  is  a  Square  ?  What  extracting  the  Square  Root  ?  What  the  Rules  ? 
What  when  a  Vulgar  Fraction  ?  What  the  Rules  in  its  Application  and 
l/*e  7  What  a  CM/*  ?  What  is,  to  extract  the  Cube  Root  ?  What  the  Rule  ? 
What  the  note,  proposition,  and  rw/e  in  its  Application  and  Use  ?  What  iho 
Rulfs  to  extract  Roots  generally  1  What  the  Note  ?  When  are  Numbers  in 
Arithmetical  Progression  1  What  form  increasing  ?  What  decreasing  ? 
How  Rawerf  ?  What  TERMS  given  ?  What  found  I  What  the  whole  nura* 


TOPICS    FOR   EXAMINATION.  Til 

her  called  1  What  Problem//^  and  rule  1  Second  and  rule  1  Third  and 
rule  ?  When  are  numbers  in  Geometrical  Progression  ?  What  called  ratio  1 
What  Problem//^  and  rule  1  What  Problem  second,  Case  first  and  rules  1 
What  Case  second,  rules  and  note  1  What  does  Alligation  teach  ?  How 
distinguished?  What  Alligation  Medial?  What  the  rule!  What  Alter- 
natel  What  the  rules  1  What  Position1*  How  many  kinds  1  What 
taught  by  Single  Position  ?  What  the  rule  ?  What  by  Double  Position  1 
What  the  rules  and  note  1  What  Permutation  1  What  Combination  ?  W  hat 
Problem  Jirst  and  rule  1  Problem  second  and  rule  1  Problem  third  and  rule  ? 
How  are  G)  indstones  sold  ?  How  are  their  Contents  found  ?  What  is  Su- 
perficial Measure  1  How  made  up  1  What  is  measured  by  it  ?  What  Case 
first  and  rule  ?  Case  second  and  rule  1  What  the  Note  ?  What  a  Triangle  1 
*How  its  Surface  measured.  How  the  Superficies  of  Joists  and  Planks  found  ? 
How  measure  irregular  Surfaces  ?  What  a  Circle  1  How  find  circumfer- 
ence if  diameter  be  given  ?  How  diameter  if  circumference  be  given  ?  How 
find  Jrea  ?  How,  if  circumference  a/one  be  given  ?  How  find  diameter  by 
the  area  ?  How  circumference  by  the  area  ?  What  is  a  Sector  ?  How  meas- 
ured ?  By  Rule  second  1  What  a  Segment  of  a  Circle  ?  How  find  its  area  1 
How  measure  a  regular  Polygon  1  How  describe  an  ellipse  or  oval  1  How 
find  its  area  ?  What  is  a  Sphere  or  Globe  1  How  find  its  area  ?  How  are 
solids  measured  1  What  is  a  Cube  1  How  measured  1  What  Case  second 
and  rule  1  What  noted  ?  What  Case  third  and  rw/e  ?  What  a  Cijlinder  1 
How  measured  ?  What  Casejifth  and  rw/e  ?  Case  sir//i  and  r?//e?  Case 


square  pyramid  ?  MOW,  ir  a  triangular  pyramid  I  Wow,  it  a  circular  pyra- 
mid, or  cone  ?  What  is  a  Globel  How  find  its  solid  content  I  What  a 
Frustum  of  a  sphere  1  How  find  its  solid  content  ?  What  is  Gauging  ? 
What  its  rw/e  and  notes  ?  How  use  the  Sliding  rule  ?  How  gauge  round 
tubs  ?  How  a  square  vessel  ?  What  the  Note  ?  How  find  a  ship's  Tonnage  ? 
What  the  A'o'e?  What  Section  fftk  and  r?//e?  Section  sixth  and  rw/e? 
How  find  the  solidity  of  Wboo1  and  Bark  ?  How  find  the  Cords  in  a  pile  of 
either?  What  the  principal  rules  in  Assessing  Taxes?  What  the  general 
Rules  in  common;  Book-keeping  ? 


VIII  ARITHMETICAL    MARKS    AND    SIGNS. 

ARITHMETICAL  MARKS  AND   SIGNS. 

=The  sign  of  equality  and  is  pronounced,  equal  to  ; 
+The  sign  of  Addition,  and  is  pronounced,  added  to  ; 
—  The  sign  of  Subtraction,  and  is  pronounced,  subtract- 
ed by. 

EXAMPLES. 

12  +  7  =  19,  twelve  added  to  seven  will  be  equal  to 
nineteen. 

23  —  8  =  15,  twenty-three  subtracted  by  eight,  equal  fif- 
teen. 

xThe  sign  of  Multiplication,  and  is  pronounced,  Multi- 
plied into  ; 
-7-The  sign  of  Division,  and  is  pronounced,  divided  by. 

EXAMPLES. 

8x7=56,  eight  multiplied  into  seven  equal  fifty-six. 
36-r-4=9,  thirty-six  divided  by  four,  equal  nine. 
Division  is  also  implied  by  the  signs  3)6(2  and  f  =2, 
six  divided  by  three,  equal  two. 

:  :  :  :  The  sign  of  Proportion,  and  is  pronounced,  is  to, 
so  is,  to. 

EXAMPLE. 
6  :  9  ::  8  :  12,  as  6  is  to  9  so  is  8  to  12. 

v/  or  *  signifies  the  Square  Root  :  thus  ^/  81  is  read,  the 
square  root  of  81  ;  or  8l¥  is  read  81  in  the  square  root. 

J.  2 

^3/  or  3  denotes  the  Cube  Root,  &c.     3    means  that  3  is 
squared,  or  to  be  multiplied,  by  itself. 

3  4 

3  means  that  3  is  to  be  cubed.  48  shows  that  48  must 
be  raised  to  the  4th  power. 

19+3x9  =  198  means  that  19  added  to  3,   and   the   sum 
multiplied  by  9,  equal  198 

~~ 

=3  shows  that  12  less  the  product  of  2  multi- 


plied  by  3,  and  divided  by  2  equal  3. 


ARITHMETIC. 


ARITHMETIC  is  the  art  and  science  of  numbers  and  has 
for  its  operation  four  fundamental  rules,  viz.  Addition, 
Subtraction,  Multiplication,  and  Division.  To  under- 
stand these,  it  is  necessary  to  have  a  perfect  knowledge 
of  our  method  of  Numeration  or  Notation. 

NOTATION 

TEACHES  to  express  numbers  by  words  or  characters. 

When  performed  by  means  of  characters  or  figures,  ten 
are  employed.  Nine  of  these  are  of  intrinsic  value  and 
are  called  digits,  or  significant  figures,  being  written  and 
named  thus  : 

1  one,  4  four,  .  7  seven, 

2  two,  5  five,  8  eight, 

3  three,  6  six,  9  nine. 

The  tenth  figure,  namely,  0,  is  called  naught  or  cipher, 
and  denotes  a  want  of  value  wherever  it  is  found. 

Besides  the  simple  value  of  the  digits,  as  noted  above, 
they  have  each  a  local  one,  which  depends  on  the  follow- 
ing principle. 

In  a  combination  of  figures,  reckoning  from  right  to 
left,  the  figure  in  the  first  place  represents  its  simple 
value  ;  that  in  the  second  place  ten  times  its  simple  val- 
ue ;  that  in  the  third  place  an  hundred  times  its  simple 
value  ;  and  so  on  ;  each  figure  acquiring  anew  a  tenfold 
value  for  every  higher  place  it  occupies.  Hence  our  sys- 
tem of  arithmetic  is  called  decimal. 

The  names  of  places  are  denominated  according  to 
their  order.  The  first  is  the  place  of  units  ;  the  second 
of  tens  ;  the  third  of  hundreds  ;  the  fourth  of  thousands  ; 
the  fifth  of  ten  thousands  ;  the  sixth  of  hundred  thou- 
sands ;  the  seventh  of  millions  ;  and  so  on.  Thus  in  the 
number  8888888  ;  8  in  the  first  place  signifies  only  eight  ; 
8  in  the  second  place  eight  tens  or  eighty  ;  8  in  the  third 
place  eight  hundred  ;  8  in  the  fourth  place  eight  thousand  ; 


10 


NOTATION. 


8  in  the  fifth  place  eighty  thousand ;  8  in  the  sixth  place 
eight  hundred  thousand  ;  8  in  the  seventh  place  eight 
millions.  The  whole  number  is  read  thus,  eight  millions, 
eight  hundred  and  eighty-eight  thousand,  eight  hundred 
and  eighty-eight. 

Though  a  cipher  has  no  value  of  itself,  yet  it  occupies 
a  place ;  and  when  set  on  the  right  hand  of  other  figures 
it  increases  their  value  in  the  same  tenfold  proportion  : 
Thus  in  the  number^SOSO ;  the  ciphers  in  the  first  and 
third  places  denote,  that,  though  no  simple  unit  or  hun- 
dreds are  reckoned,  yet  the  place  of  units  and  that  of 
hundreds  are  to  be  kept  up  to  assist  in  reckoning  the 
tens  and  thousands.  The  above  number  (8080)  is  read 
eight  thousand  and  eighty,  which,  without  the  two  ciphers, 
would  be  read  eighty-eight. 

Large  numbers  are  divided  into  periods  and  half  pe- 
riods, each  half  period  consisting  of  three  figures.  The 
name  of  the  first  period  is  units  ;  of  the  second  millions  ; 
of  the  third  billions  ;  of  the  fourth  trillions  ;  and  also, 
the  first  part  of  any  period  is  so  many  units  of  it ;  and 
the  latter  part  so  many  thousands  of  it.* 


Periods. 

Half  do. 

Figures. 


*  EXAMPLE. 
Trillions.     Billions.     Millions.        Units. 


4.       3.      2.      1. 

thou.  units.        thou.  units,     thou.  units,    thou.  c.  x.  u. 


137  462   572  329  484  617  291  337 


g        52 

I       I 
s      '£ 


>   5   § 


"§ 


NOTATION.  1  1 

NUMERATION  TABLE. 

«*>.Xxxxxxxxxxxxxxx^xxxx->xxxxxxxxxxxxxxxxxxxxxxx<$»- 

v  „  coHundreds  of  Millions. 

co  ooTens  of  Millions. 
co  GO  <iMillions.  ^ 

CD  QO  -5  OiHundreds  of  Thousands.  * 
CD  QO  <i  a  CttTens  of  Thousands.  ^ 

o  GO  <i  o  0*  ^Thousands. 
co  OD  <?  cs  CN  £*  csHundreds. 
\      o  QD  <!  o  CK  rf*  co  ioTe-ns. 
\  cooo<!0:^^coto  H-Units.  \ 


\ 


To  express  in  figures  Numbers  which  exceed  Nine. 
RULE.  —  Write  down  ciphers  to  so  many  places  as  are 
named  in  the  given  number  ;  then,  beginning  at  the  left, 
observe  at  each  place  what  significant  figure  is  named, 
and,  taking  away  the  cipher,  write  the  significant  figure 
in  its  place  ;  and  thus  proceed  with  each  place  till  you 
come  to  the  place  of  units. 

EXAMPLES. 

Twenty-five, 
One  hundred, 

Three  thousand  and  fifteen, 
Eight  hundred  and  twelve  thousand, 
Thirty-one  thousand,  two  hundred  and  six, 
Six  millions,  seven  thousand  and  eight, 
One  hundred  and  one  millions,  fourteen  thousand  and 
fourteen. 

To  read  NUMBERS. 

RULE.  —  First  numerate,  from  the  right  to  the  left  hand, 
each  figure,  in  its  proper  place,  by  saying,  units,  tens^ 
hundreds,  &c.,  as  in  the  Numeration  Table.  Then,  to 
the  simple  value  of  each  figure,  join  the  name  of  its 
place,  beginning  at  the  left  hand,  and  reading  to  the  right. 

EXAMPLES. 
64, 
396, 
4015, 
76920, 
104080, 
5300648, 


SIMPLE    ADDITION. 

NOTE.—  The  pupil  should  be  accustomed,  in  each  Example,  in  the  follow- 
ing Rules,  to  read  correctly  not  only  every  answer,  but  every  line  of  num- 
bers in  his  sum. 


ADDITION. 

ADDITION  in  Arithmetic  is  the  uniting  or  joining  to- 
gether of  two  or  more  numbers. 

SIMPLE  ADDITION  is  the  collecting  of  several  numbers, 
of  the  same  denomination  into  one  sum  ;  as,  4  yards  and 
6  yards,  expressed  in  one  sum,  are  10  yards. 
Addition  and  Subtraction  Table. 

<>XXXXXXXXXXXXXX^XXXXXXXXXXXXX-XXXXX^ 

•  j    1|  2|  3|  4|  5[  6|  7|  8| 


2|  4|  5|  6[  7|  8|  9|10|llj]2|13|14 

\   3|  5|  6|'7|  8|  9|IOJTTfl2|13|14|15  * 

$"4|  6|  7|"8r9|10jil|ia|13]l4il5fi6^ 

*  5|  7|  8[  9|10|IT|12|73|14|15|16|17  * 
}   6[  8|  9|10|ll|12fl3|14|15|I6|17|18  * 

*  7|  9|10|I1|12|T3|14|I5|16|17|18|19  * 


\   9|ll|l2|l'3iT4n5[i6|i7|18|19|20|2i  \ 
\  10|12|  I3|  14|  15|  I6|  17|  18|  I9|20|21  122  ^ 

<^xxxxxxxxxxxxxx^-x~~xxx~xxxx^ 

When  you  would  add  two  numbers,  seek  one  of  them 
in  the  left  hand  column,  and  the  other  in  the  top  line  ; 
and  in  the  common  angle  of  meeting,  or  at  the  right  hand 
of  the  first,  and  under  the  second,  you  will  find  the  sum  ; 
as,  6  and  9  are  15;  and  so  of  any  others. 

When  you  would  subtract,  seek,  in  the  left  hand  col- 
umn, the  number  to  be  subtracted  from  the  greater  ;  then 
run  your  eye  along,  in  the  same  line,  towards  the  right 
hand,  till  you  find  the  number  from  which  the  other  is  to 
be  taken  ;  and  exactly  over  this  last,  in  the  top  line,  you 
will  find  the  difference  ;  as,  6  from  15,  and  there  remain 
9  ;  and  so  of  any  others. 

SIMPLE  ADDITION. 

RULE.  —  Write  the  numbers,  units  under  units,  tens  un- 
der tens,  &c.  and  draw  a  line  under  the  whole.  Add  up 
the  unit  column,  and  if  the  sum  be  less  than  ten,  write 


SIMPLE    ADDITION. 


13 


it  under  the  column ;  if  it  be  ten  or  any  number  of  tens, 
write  a  cipher;  if  there  be  an  excess  over  ten  or  tenst 
write  down  this  excess,  and  carry  as  many  units  to  the 
next  column,  as  there  are  tens  ;  and  thus  proceed  with 
each  remaining  column,  writing  the  whole  sum  under 
the  last. 

PROOF. — Draw  a  line  below  the  upper  number,  and  add 
the  remaining  numbers  as  shown  in  the  rule  ;  add  the  sum 
thus  found  and  the  upper  number  together,  and  if  the  sum 
be  equal  to  the  first  addition,  the  work  is  right.  Or, 
begin  at  the  top  number,  add  downwards,  and  carry  as 
before  ;  if  the  two  sums  come  alike,  the  work  is  probably 
right 

EXAMPLES. 

2. 

567842 
143469 
782107 
695213 
203169 


14786 


4. 


s=3  .  o 


S3    3 


98765432 
46532815 
90054061 
00103 


327 

3  2 


15000 


9132051 
460109 


65400123 
86194217 
28103019 
17631042 
98765208 
37849000 
54001605 


B 


14  S13IPLE    ADDITION. 

6.  7.  8. 

Miles.  Leagues.     Years. 

4734746  46434733  347312484 

3474352  74265374  368126312 

4634324  52652754  758612691 

7369138  35374265  731674591 

3543468  74447352  323473276 

4733246  47345264  471266198 

4743447  74167574  323634712 

3752612  43526526  271254712 

7426984  38573452  312844795 


APPLICATION. 

1.  What  is  the  sum  of  37,  509,  7126,  17630,  and 
459273  yards  1 

2.  Required,  the  sum  of  3579,  41,  96120,  725,  11, 
1820,  5,  and  720139  bushels. 

3.  What  is   the   sum    of  2591,  720396,   14,  259,  6, 
370214,  9740,  53,  1692,  and  137  dollars  ? 

4.  How  many  days  are  in  the  12  calendar  months,  in  a 
leap  year  ? 

5.  A  person  dying  left  to  his  widow  1500  dollars,  to  his 
eldest  son  30500,  to  each  of  his  other  two  3406  ;  also  2700 
to  each  of  his  three  daughters,  besides  751  dollars  in  other 
small  legacies  ;  what  did  his  estate  amount  to  ? 

6.  If  the  distance  from  Hallowell  to  Portland  be   fifty 
six  miles,  thence  to  Portsmouth  fifty-four  miles,  thence  to 
Boston  sixty-four  Iniles,  thence  to  Hartford   ninety-eight 
miles,  thence  to  New- York  one  hundred  and  eleven  miles, 
thence  to  Philadelphia  ninety  miles,  thence  to  Baltimore 
ninety-nine  miles,  and  thence  to  Washington  thirty-eight 
miles  ;  what  is  the  whole  distance  between  Hallowell  and 
the  city  of  Washington  1 

7.  John,  James,  and  Paul  counting  their  prize-money, 
John  had  one  thousand,  three  hundred  and  seventy-five 
dollars  ;  James  had  just  three  times  as  much  as  John: 
and  Paul  had  just  as  much  as  both  the  others  ;  pray  how 
many  dollars  had  Paul  ? 


SIMPLE    SUBTRACTION.  15 

SIMPLE  SUBTRACTION. 

SUBTRACTION  is  finding  the  difference  of  two  numbers, 
by  taking  the  less  from  the  greater.  It  is  simple  subtrac- 
tion if  the  numbers  are  of  one  denomination  ;  as,  5  feet 
taken  from  8  feet,  will  leave  3  feet. 

The  greater  number  is  called  the  minuend,  or  substra- 
tum; the  less,  the  subtrahend;  and  the  number  found  by 
the  operation,  the  difference,  or  remainder. 

RULE. — Write  the  less  number  under  the  greater,  plac- 
ing units  under  units,  tens  under  tens,  &c.  and  draw  a 
line  under  them.  Begin  at  the  right,  and  take  each  figure 
in  the  subtrahend  from  its  corresponding  one  in  the  min- 
uend, setting  down  the  remainder  straight  under  it  below 
the  line.  If  the  lower  figure  be  greater  than  the  one 
above  it,  add  ten  to  the  upper  figure,  from  which  sum  take 
the  lower,  and  set  down  the  remainder,  carrying  one  to 
the  next  lower  figure  ;  and  thus  proceed  until  the  whole 
is  finished. 

PROOF. — Add  the  remainder  to  the  subtraJiend,  and  if 
the  sum  be  equal  to  the  minuend,  the  work  is  right. 

EXAMPLES. 

1.  2.  3. 

From  67216  the  minuend,         46132941  71290 

Take  43792  the  subtrahend.      17316257  46172 


23424  Remainder. 


67216  Proof. 

5.  6.  *  7. 

87652176  100000  200000 

9107215  65321  99999 


8.  9. 

10000  917144043605 

1  40600S32164 


16                             SIMPLE  MULTIPLICATION. 

10.  11. 

100200300400  1 0000000 

98087076065  9999991 


APPLICATION. 

1.  From  360418  tons,  take  293752. 

2.  From  100046  acres,  take  10009. 

3.  What  is  the  difference  between  1735,  and  1897348 
hours  1 

4.  How  much  do  540312  days  exceed  7953  ? 

5.  How  much  are  30491  gallons  less  than   57321469  ? 

6.  If  the  distance  from  Hallowell  to  Savannah,  through 
Washington,  be  1268  miles  and  that  from   Washington 
to  Savannah,  658  miles  ;  how  far  is  Washington  from 
Hallowell  ? 

7.  From  Hallowell  to  the  city  of   New-  York  is  383 
miles.     Now,  if  a  man  should  travel  10  days  from  Hal- 
lowell towards  New-York,  at  the  rate  of  thirty-six  miles 
each  day  ;  how  far  would  he  then  be  from  that  city  ? 

8.  If  a  farmer  kills  six  hogs,  which  weigh  two  hundred 
and  fifty-four,  one  hundred  and  ninety-seven,  two  hun- 
dred and  sixteen,  two  hundred  and  forty-nine,  three  hun- 
dred and  twelve,  and  three  hundred  and  sixty-three,  and 
markets  one  thousand  weight  of   pork  ;    what  quantity 
does  he  reserve  for  his  own  use  1 


SIMPLE  MULTIPLICATION. 

MULTIPLICATION  is  finding  the  amount  of  any  given 
number,  by  repeating  it  any  proposed  number  of  times  ; 
as,  4  times  7  are  28. 

The  number  to  be  multiplied  is  called  the  multiplicand. 

The  number  which  multiplies  is  called  the  multiplier. 

The  number  arising  from  the  operation  is  called  the 
product. 

The  multiplicand  and  multiplier  are  called/actors  ;  and 
if  these  are  of  one  denomination  it  is  called  Simple 
Multiplication. 


SIMPLE    MULTIPLICATION.  17 

MULTIPLICATION  AND  DIVISION  TABLE. 

}  ^V[^^^^\^^\^^\^^\^  \ 
\  2|  4|  6|  8|10|12|14|16|  18|  20|  22 1  24$ 
\  3|  6|  9|12|15|18J21|24|  27["3QT33|  36  \ 
*  4I  8|12|16|20|24|28|32|  36|  4-0 1  44j  48  t 
$  5|10|15|20j25|30i35|40|'45|  50|  55 1  60  } 
$lpp3J24|30|36[42|48|  54|~6Q|  66 1  72* 
\  7|14|21|28|35i42|49|56J^3|  70|  77|  84  * 
x  8|16|24|32i40|48J56|64|  72|  80[  88 1  96  $ 
N  9|18i27j36J45{54J63|72|  81|  90 1  99|108  i 
i  10i20|30[40;50i60|70|80|  9Q|100|UO|12Q  | 
iri|22i^44155|66j77p^9JI^^  ^ 

^  12|24|36|48|60l,72|84i96ilOSjl20(132|144  $ 
<^~~ „„„„, +„„„ — +~^ 

USE  of  the  TABLE  in  MULTIPLICATION. 
Find  the  multiplier  in  the  left    hand  column,  and  the 
multiplicand  in  the  uppermost  line ;  and   the  product  is 
in  the  common  angle  of  meeting,   or  against  the  multi- 
plier, and  under  the  multiplicand. 

To  use  the  ahove  Table  in  Division,  seek  your  divisor 
in  the  left  hand  column  ;  then  run  your  eye  along  the 
line,  to  the  right  hand,  till  you  come  to  your  dividend  ; 
and  the  figure  in  the  top  line,  of  the  same  column,  will 
be  the  quotient,  or  number  of  times  the  divisor  is  con-' 
tain'ed  in  the  dividend. 

CASE  I.—^When  the  multiplier  is  not  more  than  twelve. 
RULE.— Multiply  each  figure  in  the  multiplicand  by  the 
multiplier,  beginning  at  the  right  hand  side,  and  setting 
down  the  whole  of  such  products  as  are  less  than  ten  ; 
but  for  such  as  are  just  equal  to  a  certain  number  of  tens, 
write  down  0,  and  carry  ]  for  each  ten  to  the  next  pro^ 
duct ;  and  for  such  as  exceed  a  certain  number  of  tens, 
set  down  the  excess,  and  carry  for  the  tens  as  before. 

EXAMPLES. 

1.  What  number  is  equal  to  4  times  365  ? 
Thus  2. 

365  Multiplicand.  5124167 

4  Multiplier.  3 

Ana.    1460  Product. 
B3 


18  SIMPLE    MULTIPLICATION. 

3.  4. 

42179416        74216 
5  2 


11. 

567295 
12 


CASE  II. — When  the  multiplier  consists  of  several  figures. 

RULE. — Set  the  multiplier  under  the  multiplicand,  so 
that  units  may  be  under  units,  tens  under  tens,  <fec.  then 
find  the  product  for  each  figure  in  it,  as  in  the  first  caset 
not  regarding  in  what  order  the  lines  are  found,  provided 
the  first  figure  in  each  stand  straight  below  its  respective 
multiplier.  Add  all  the  lines  of  products  together  in  the 
same  order  as  they  stand,  and  the  sum  will  be  the  whole 
product  required. 

PROOF. — Make  the  former  multiplicand  the  multiplier, 
and  the  multiplier  the  multiplicand,  and  proceed  as  be- 
fore ;  and  the  new  product  will  be  the  same  as  before, 
when  the  work  in  both  is  right.  Or,  add  together  the  fig* 
ures  first  of  one  factor,  and  then  of  the  other,  casting 
out  all  the  nines  in  the  sums  of  each,  as  often  as  they 
amount  to  9.  Multiply  the  two  remainders,  if  any,  to- 
gether, and  the  nines  cast  out  of  their  product,  will  leave 
the  same  remainder  as  the  nines  cast  out  of  the  answer, 
when  the  work  is  right.  The  first  remainder  may  be  set 
at  the  left  side  of  the  cross,  or  X  ;  the  second  at  the  right ; 
that  arising  from  their  product  at  the  top  ;  and  that  aris- 
ing from  the  answer  at  the  bottom ;  if  the  answer  ba 
right,  the  top  and  bottom  figures  will  be  alike* 


SIMPLE    MULTIPLICATION.  19 

EXAMPLES. 

I.  329  excess  of  9s,  5 

4271  Multiplicand.  4271      do.        do.  5 
329  Multiplier. 

329 

38439  2303 

8542  658 

12813  1316 

1405159  Product.     Proof  1405159  excess  of  9s,  7 

7 
or  5x5 

7 

2.  3.  4. 

691861          129186  281216 

26  98  978 


17988386        12660228         275029248 


5.  6.  7. 

181281          281691         264648436 
763          76286          3639604 


138317403      21489079626    963215506259344 


CONTRACTIONS  IN  MULTIPLICATION. 

CASE  I. — When  there  are  ciphers  at  the,  right  hand  of  one 

or  both  of  the  factors. 

RULE. — Proceed  as  before,  neglecting  the  ciphers,  and 
to  the  right  hand  of  the  product,  place  as  many  ciphers 
as  are  in  both  the  numbers. 

EXAMPLES. 

1.                            2.  3. 

27600        180120  27640 

48000         48100  20 


2208 
1104 


1 324800000      8663772000 


20  SIMPLE    MULTIPLICATION. 

CASE  II. —  Wfien  the  multiplier  is   the  product  of  two   or 

more  numbers. 

RULE. — Multiply  once  successively  by  each  of  those 
numbers  instead  of  using  the  whole  multiplier  at  once, 

EXAMPLES. 

1.  % 

Multiply  7629  by  63  74639  by  72 

7x9=63  7 

3 
46217  by  96 


4. 


Prod.  480627  37692  by  132 

APPLICATION. 

1.  What  will  37  horses  for  shipping  come  to,   at  52 
dollars  per  head  1  Ans.  1924  dols. 

2.  What  will  587  firkins  of  butter  come  to,    at  7  dol- 
lars per  firkin  ?  Ans.  4109  dols. 

3.  What  will  367  acres  of  land  cost,  at  13  dollars  per 
acre  ?  Ans.  4771  dols. 

4.  If  a  barrel  of  pork  cost  18  dollars  what   will  857 
barrels  be  worth  ?  Ans.  15426  dols. 

5.  What  will  be  the  worth  of   924  tons   of    potash,  if 
one  ton  sell  for  95  dollars  1  Ans.  87780  dols. 

6.  A  merchant  having  traded  ten  years,  found  he   was 
worth  13000  pounds.     His    books  showed   that  the   last 
three  years  he  had  cleared  873  pounds  a  year ;  the  three 
preceding  but  586  pounds  a  year  ;    and  before   that  but 
364  pounds  a  year.     With  what  sum  did   he  begin  busi- 
ness ?  Ans.  7167  pounds. 

7.  Trajan's  bridge  over  the  Danube  is  said  to  have  had 
twenty  piers  to  support  the  arches,  every   pier  being  60 
feet  thick,  and  some  of  them  .150  feet  above    the   bed   of 
the  river  ;  they  were  also  170  feet  asunder.     Pray,  how 
wide  was  the  river  in  that  place  1  and  how  much  did  this 
bridge  exceed  in  length  that  at  Westminster,  in  England, 
which  is  about  1200  feet  from  shore  to  shore,  and  is  sup- 
ported by  11  piers,  making  the  number  of  arches  12  I 

i        f  4770  feet  wide,  and  3570  feet  lon~ 
5*  (      ger  than  Westminster  bridge, 


SIMPLE    DIVISION.  21 

8  In  the  partition  of  lands  in  a  certain  settlement,  A. 
had  757  acres  allotted  to  him  ;  B.  2104  ;  C.  16410  ;  D. 
12881;  E.  11008;  F.  9813;  H.  13800;  and  I.  8818 
acres.  Now  as  the  above  allotments  want  416  acres  to 
make  them  just  one  fifth  of  the  whole,  how  many  acres 
did  the  settlement  contain  T  Ans.  380035  acres. 


DIVISION. 

DIVISION  shows  how  often  one  number  is  contained  in 
another  :  as  24  divided  by  6,  produce  4  in  the  quotient  ; 
that  is,  6  are  contained  4  times  in  24. 

The  number  to  be  divided  is  called  the  dividend. 

The  number  by  which  we  divide  is  called  the  divisor. 

The  number  of  times  the  dividend  contains  the  divisor 
is  called  the  quotient. 

The  remainder,  if  there  be  any,  will  be  less  than  the 
divisor. 

It  is  called  Simple  Division,  if  the  dividend  and  divisor 
have  but  one  and  the  like  name. 

RULE.  —  On  the  right  and  left  of  the  dividend  draw  a 
curved  line,  and  write  the  divisor  on  the  left,  and  the  quo- 
tient as  it  arises  on  the  right  hand.  Assume  as  many  fig- 
ures on  the  left  hand  of  the  dividend  as  contain  the  divi- 
sor once,  or  more,  and  place  the  number  in  the  quotient. 
Multiply  the  divisor  by  the  quotient  figure,  and  set  the 
product  under  the  assumed  part  of  the  dividend  ;  subtract 
it,  and  to  the  remainder  bring  down  the  next  figure  of  the 
dividend  ;  which  number  divide  as  before,  and  thus  pro- 
ceed until  the  whole  is  divided. 

NOTE  1.  —  If  after  a  figure  is  brought  down,  the  number 
be  less  than  the  divisor,  place  a  cipher,  in  the  quotient,  and 
bring  down  the  next  figure  of  the  dividend. 

NOTE  2.  —  Remember  that  the  products  of  the  divisor 
and  the  several  quotient  figures,  must  always  be  less  than 
the  parts  of  the  dividend  under  which  they  are  set,  unless 
they  chance  to  be  just  the  same  numbers  ;  and  that  every 
remainder  must  be  less  than  the  divisor. 

PROOF.  —  To  the  product  of  the  divisor  and  quotient, 
add  the  remainder,  which  sum  will  be  equal  to  the  divi- 
dend, if  the  work  is  right. 


SIMPLE    DIVISION. 


Or,  cast  the  nines  out  of  the  divisor,  and  place  the  ex- 
cess or  remainder  on  the  leftside  of  a  cross,  or  X  ;  do  the 
same  with  the  quotient,  and  place  the  excess  on  the  right 
hand  ;  multiply  these  two  figures  together,  add  their  pro- 
duct to  the  remainder  of  the  divisor,  if  there  be  any,  cast 
out  the  nines  in  the  sum,  and  set  the  excess  at  the  top  of 
the  cross;  cast  the  nines  Out  of  the  dividend,  and  place 
the  excess  at  the  bottom  ;  then,  if  the  top^  and  bottom 
figures  are  alike,  the  work  is  right. 

EXAMPLES. 
I  2 

Divis.    Divid.     Quot. 
48)7641312(159194 

48  48  '  15 


Divis.  Divid.     Quot. 
5)172164(34432f 


284 


1273552 


240        636776 


441 
432 


7641312  Proof. 


22 
20 

21 
20 


93       Proof  by  excess 

48  of  nines. 

6 

451  3x2 

432  6 

192 
192 


16 
15 

14 
10 


Proof. 

3 

5x7 
3 

5x7=35 
add  4  rem. 

39 


4  Remainder. 


NOTE. — If  there  be  no  remainder,  the  quotient  is  the 
perfect  answer  to  the  question  ;  but  if  there  is,  to  com- 
plete the  quotient,  put  the  remainder  at  the  end  of  it,  and 
the  divisor  below  it,  drawing  a  line  between  the  two. 

Quotient.         Rem. 
by     29         Ans.  5296     and      14 
63  4780  7 

763  181281  0 

232  48340  7 

400  11695  216 

3215  0 

2359  1255 

6000  4711 


3.  Divide  153593 

4. 

301147 

5. 

138317403 

6. 

11214887 

7. 

4678216 

8. 

1030603615 

9. 

4917968967 

10, 

210634711 

SIMPLE    DIVISION.  23 

CONTRACTIONS. 

CASE  I. — When  there  are  ciphers  at  the  right  hand  of 
the  divisor,  cut  them  off;  likewise  cut  off  the  same  num- 
ber of  digits  from  the  right  hand  of  the  dividend  ;  then 
divide  as  usual,  and  to  the  remainder  annex  the  digits 
cut  off  from  the  dividend. 

EXAMPLES. 

1.  2. 

342,00)6792,16(19  135,000)27619,413( 

342 

3372 
3078 


29416  Remainder.  79413  Rem. 

CASE   II. —  When  the  divisor  is  any  number  not  exceed- 
ing 12. 

RULE. — First  seek  how  often  the  divisor  can  be  had  in 
the  first  figure,  or  figures,  of  the  dividend  ;  put  the  result 
under  the  dividend  ;  multiply  this  quotient  figure  and  the 
divisor  together  ;  mentally  subtract  their  product  from,  the 
part  of  the  dividend  taken  ;  what  remains  call  so  many 
tens,  which  place,  in  idea,  before  the  next  figure  of  the 
dividend  for  a  new  dividual ;  and  so  proceed  through  the' 
whole  dividend.  When  in  subtracting,  nothing  remains, 
take  the  next  figure  ;  if  that  be  less  than  the  divisor, 
take  the  next  two,  and  place  a  cipher  under  the  first. 

1.  2. 

6)7241324(  5)172164( 


Quotient  1206887f  Rem.         Quotient  34432f  Rem. 

3.  Divide  3764592     by     7 

4.  527684     8 

5.  1410217     9 

6.  612948    11 

7.  317926    12 


24  SIMPLE    DIVISION. 

CASE  III. — If  the  divisor  be  a  product  of  two  or  more 
numbers,  divide  continually  by  those  numbers  instead  of 
the  whole  at  once. 

EXAMPLES. 

1.  2. 

Divide  7621460  by  16  4792161  by  48 

4)7621460    '  6)4792161 


4)1905365  8)          —3 

Quo.  476341—4  Rem.  99836—5x6+3=33 

NOTE. — It  sometimes  happens  that  there  is  a  remainder 
to  each  of  the  quotients,  and  neither  of  them  the  true  one 
but  the  true  remainder  maybe  found  by  the  following  rule. 

RULE. — Multiply  the  last  remainder  by  the  last  divisor 
but  one,  and  to  the  product  add  the  preceding  remainder ; 
multiply  this  sum  by  the  next  preceding  divisor,  and  to 
this  product  add  the  next  preceding  remainder,  and  so 
on  until  all  the  remainders  and  divisors  are  used ;  and 
the  last  sum  will  be  the  true  remainder. 

3.  4. 

Divide    6421671  by  448        Divide  27162  by  62 
8x8x7=448        8)6421671 

8)802708—7 

7)100338—4 
Quotient.     14334—4x8+7=39  Remainder. 

APPLICATION. 

1.  Divide  3656  dollars  equally  among  8  men. 

Ans.  457  dols.  to  each. 

2.  There  are  124  men  who  have  372  dollars  among 
them  ;  how  much  is  one  man's  share,  if  it  be  divided 
equally  ?  Ans.  3  dols. 

3.  If  I  wish  to  perform  a  journey  of  3264  miles  in  136 
days,  how  far  must  I  travel  each  day  to  complete  it  ? 

Ans.  24  miles* 


PRACTICAL 'QUESTIONS.  25 

4.  A  payment  of  1272  dollars  was  made  by  a  number 
of  men,  each  of  whom  paid  3  dollars  ;  how   many    men 
were  there  T  Ans.  424. 

5.  I  would  plant  2072  trees,  in  14  rows,  25  feet  asun- 
der; how  long  must  the  grove  be .  ?         Ans.  3675  feet. 

6.  Divide  1000  dollars,  between  A,  B,  and  C,  and  give 
A  129  more  than  B,  and  B  178  less  than  C. 

Ans.  360  dots.  A's,  231  B's,  and  409  C's. 

7.  Part  1500  acres  of  land   between   Saul,   Seth,   and 
Silas;  and  give  Seth  72  more  than  Saul,  and   Silas   112 
more  than  Seth.  .        (  414f  Saul's  share,  486f 

3'  \  Seth's  and  598f  Silas's. 

8.  A  brigade  of  horse  consisting  of  384  men,  is  to  be 
formed  into  a  column,   having  32   men  in    front  ;    how 
many  ranks  will  there  be  ?  Ans.  12. 

9.  In  order  to  raise  a  joint  stock  of  10,000  dols.  L,  M, 
and  N,  together,  subscribe  8500,  and  O  the   rest.     Now, 
M  and  N  are  known  together  to  have  set  their   hands  to 
6050,  and  N  has  been  heard  to  say  that  he  had  undertaken 
for  420  more  than  M.  What  did  each  proprietor  advance? 

Ans.  L  2450,  M  2815,  N  3235,  and  O  1500. 


PRACTICAL  QUESTIONS, 

UNDER  THE  PRECEDING  RULES. 

1.  Add  fourteen  thousand,  five  hundred  and  nine  ;  one 
thousand,  nine  hundred  and  twenty-one  ;  six  hundred  and 
twenty  thousand, three  hundred  and  forty-seven;  and  live 
million,  twenty-three  thousand,   arid    nineteen,   together. 

Ans.  5659796  sum. 

2.  What  is  the  sum  of  76129+54216+39127+62357 
+  514026?  Ans.  745855. 

3.  What  is  the   difference  between   four  million  two 
hundred  and  ten  thousand  and  twelve;  and  six  hundred 
and  fifty-nine  thousand  seven  hundred  and  ninety-seven  1 

Ans.  3550215. 

4.  Take  nine  hundred  and  one  thousand  and  fifteen, 
from  one  million  one  thousand  one  hundred  and  one  ? 

Ans.  100086. 

5.  A  farm  of  460  acres  is  let  for  2  dollars  per  acre  ; 
bow  much  does  the  rent  amount  to  ?        Ans.  920  dols. 

C 


26  PRACTICAL    QUESTIONS. 

6.  If  a  man's  income  be  6  dollars  a  day,  how  much 
does  it  amount  to  in  a  year,  allowing  365  days  in  a  year  ? 

Ans.  2190  dollars. 

7.  What  is  the  product  of  376x54  ?        Ans.  20304. 

8.  64  men  have  17280  dollars  divided  equally  among 
them  ;  what  is  each  man's  part  ?         Ans.  270  dollars. 

9.  Multiply  three  hundred  and  seventy-eight  thousand 
and  five  hundred,  by  thirty-four.  Ans.  12869000  product. 

10.  What  is  the  third  part  of  3669  1         Ans.  1223. 

11.  Divide  6764  by  19.  Ans.  356  quotient. 

12.  What  number  must  be  added  to  764  to  make  it 
1256  ?  Ans.  492. 

13.  By  what  number  must  I  multiply  67,  that  the  pro- 
duct may  be  871  ?  Ans.  13. 

14.  There  are  two  numbers  whose  difference  is  796, 
the  greater  number  is  4320  ;  I  demand  the  less. 

Ans.  3524. 

,    15.  Supposing  a  man  to  have  been  born  in  the  year 
1762  ;  how  old  was  he  in  1806  ?  Ans.  44. 

16.  Suppose  a  man  to  have  been  78  years  old  in  the 
year  1806  ;  in  what  year  was  he  born  ?         Ans.  1728. 

17.  What  will  12  tons  of  hay  come  to  at  27  dollars 
per  ton  ?  Ans.  324  dollars. 

18.  What  will  750  barrels  of  beef  come  to  at  11  dol- 
lars per  barrel ;  and  what  will  the  profits  amount  to  in 
selling  it,  if  I  clear  3  dollars  on  each  barrel  ? 

A        (  8250  dollars  amount. 
'*  \  2250  dollars  profit. 

19.  There  is  a  town  which  contains  290  houses,  and 
each  house  6  inhabitants ;    how  many  inhabitants  are 
there  in  that  town  ?  Ans.  1740. 

20.  A  prize  of  48726  dollars  is  owned  by  270  men  ; 
what  is  each  man's  share  ?  Ans.  ISOf-J  dollars. 

21.  If  12  bundles  of  wheat  produce  I  bushel,  how  many 
bushels  will  4764  bundles  produce  ?    Ans.  397  bushels. 

22.  Borrowed  of  AJ  12  sums  of  money,  each  250  dol- 
lars ;  paid  him  at  one  time  97  dollars,  and  at  another  35 ; 
the  balance  1  am  to  pay  him  in  six  equal  payments  ;  what 
is  one  of  those  payments  7  Ans.  478  dollars. 


MONEY,    WEIGHTS,    MEASURES,    &C.  2? 

TABLES 

OF  MONEY,  WEIGHTS,  AND  MEASURES. 
1.  MONEY,* 

4  Farthings  make  one  penny  ;  qr.  d.  denote  farthings 

and  pence  respectively. 

12  Pence  make  one  shilling      -     -     5     -     -     Shilling. 
20  Shillings  1  pound         .--,£--         Pound. 
£  Is  one  farthing,  or  one  fourth ;  %  is  one  halfpenny, 
or  one  half;  £  three  farthings,  or  three  fourths. 

2.  TROY  WEIGHT. 

24  Grains  make  one  pennyweight,  marked  grs.   dwt. 
20  Pennyweights  -     1  Ounce,         -       -     -     oz. 

12  Ounces  -     -     1  Pound,     -      -     fo-  or  Ib. 

By  this  weight  are  weighed  jewels,  gold,  silver,  electu- 
aries and  liquors, 

3.  APOTHECARIES'  WEIGHT. 

80  Grains  make    -     1  Scruple,  marked  gr.  J),orser. 
3  Scruples    -    -       1  Dram,      -      -    -         3,  or  rfr. 
8  Drarns    -    -    -     1  Ounce,       -      -    -      Z,  or  oz. 
12  Ounces    -    -    -    1  Pound,     -    -    -    -    jfc.  or  Ib. 
Apothecaries   use  this  weight  in  compounding  their 
medicines,  but  they  buy  and  sell  their  drugs  by  Avoirdu- 
pois weight. 

*  Sterling-  money  was,  fornrafy,  of  the  same  value  in  all  the 
Colonies  of  North- America.  ™y  reason,  however,  of  the  emis- 
sion  of  paper  money  by  the  Legislatures  of  those  Colonies, 
which  afterwards  depreciated,  the  Spanish  dollar  came  to  be 
reckoned,  in  different  Colonies,  at  a  higher  or  lower  value,  ac- 
cordingly to  the  less  or  greater  depreciation  of  their  paper  cur- 
rencies. Still,  though  the  pound  was  valued  accordingly  to  this 
paper  medium,  it  was,  in  every  Colony,  reckoned  at  twenty  shil- 
lings, as  in  England.  Thus,  a  Spanish  dollar  being-  worth  4*.  6d. 
in  England,  became,  in  Georgia  and  South- Carol  in  a,  where  the 
depreciation  of  the  paper  was  least,  worth  4s.  8d. ;  in  Canada 
and  Nova-Scotia,  where  it  was  somewhat  greater,  5s. ;  in  New- 
England,  Virginia,  Kentucky,  and  Tennessee,  6s  ;  in  New- Jer- 
sey, Pennsylvania,  Delaware  and  Maryland,  7s.  6d.  ;  and  in 
New-York,  and  North-Carolina,  8s. 


WEIGHTS,    MEASURES,    &C. 


4.  AVOIRDUPOIS  WEIGHT. 


16  Drams  make 
16  Ounces 
28  Pounds* 

4  Quarters     - 
20  Hundred  wt. 


1  Ounce,  marked,  -  dr.  oz. 
1  Pound,  -  Jk  or  Tb. 

1  Quarter,  -  -  -  qr. 
1  Hundred  weight.  -  cwt. 
1  Ton,  ...  T. 


By  a  late  law  of  this  State,  Soffomake  a  qr. 

By  this  weight  are  weighed  all  things  of  a  coarse  na- 
ture ;  such  as  leather,  cheese,  grocery  wares,  bread,  and 
all  metals  except  gold  and  silver.  It  is  our  common 
steelyard  weight. t 

NOTE.— 6760  grains— 1ft.  Troy;  7000  grains— 1ft. 
Avoirdupois  ;  therefore  the  weight  of  a  pound  Troy,  is  to 
a  pound  Avoirdupois  as  5760  to  7000,  or  as  144  to  175. 

5.  CLOTH  MEASURE. 

4  Nails,  or  9  inches,  make  1   Quarter,  marked,  na.  qr, 

4  Quarters     -                   -     1  Yard,  -     yd 
3  Quarters         -         -       -  1  Ell  Flemish,  -       -  E.  Fl 

5  Quarters     -         -       -      1  Ell  English,       -  E.  E 

6  Quarters         -  1  Ell  French,    -       -  E.  Fr\ 

NOTE. — We  buy  Scotch  and  Irish  linens  by  our  Amer- 
ican yard,  and  Dutch  linens  by  the  eli  Flemish  ;  but  we 
sell  them  here  by  the  same  measure,  the  yard. 

*  Most  of  ^e  merchants  and  traders  in  the  United  States,  now 
call  25  Ibs.  onlyj^  quarter  of  a  cwt. 


f  A  Firkin  of  Butter  is  56  Ib. 

A  firkin  of  Soap  64 

A  Barrel  of  Beef  220 

Pork  220 

Potashes  400 

Soap  256 

Butter  224 

Flour  196 

Gunpowder  112 

Raisins  112 
Fish          30  gallons 


A  Quintal  of  Fish  112lb. 

of  Iron  14 

n  of  Train  Oil  7£ 

20  Thing-s  make  1  Score. 

12       do.  1  Dozen. 

12  Dozen  1  Gross. 

144     do.  1  Great  Gross. 

A  Quire  of  Paper       24  Sheets. 

A  Ream  of  Paper      20  Quires. 

A  Bale  of  Paper        10  Reams. 

A  Roll  of  Parchment  60  Skins. 


Hoops  and  Staves  are  now  reckoned   five  scores  to  the  hun- 
dred in  this  State,  by  a  late  law. 
A  ton  in  weight  for  Ships  is  2000  Ib. 
A  ton  for  goods,  boxes,  cases,  &c.  is  40  cubic  feet. 


MEASURES.  29 

6.  LONG  MEASURE. 

3  Barley  Corns  make  1  Inch,  marked  Bar.  In. 

12  Inches  1  Foot,  -  Ft. 

3  Feet,  -  -  1  Yard,  -  -  Yd. 

5£  Yards,  or  16£  Feet  1  Pole,  Rod  or  Perch,  Rod. 

40  Poles  or  Rods  1  Furlong,  -  -  -  Fur. 

8  Furlongs     -      -       -  1  Mile,     -  Mi. 

3  Miles    -  1  League,    -  Lea. 
60  Geographical,  or    \  *  DpOTPP  rw 

69£  Statute  Miles          }        1>e^ree'  JJe^' 

360  Degrees  make  the  circumference  of  the  Earth. 

By  this  measure  distances  are  measured. 
66  feet,  or  4  rods,  make  a  Gunter's  chain,  containing 

100,  links,  each  of  which  is  7T9/o  inches. 
6  feet  make  a  fathom,  in  measuring  depths. 
5  feet  make  a  geometrical  pace. 

4  inches  make   a  hand,    in  measuring  the  height  of 
horses.     6ft.  44in=a  French  toise. 

1  French  post=%  Fr.  Leagues=5T5^  Eng.  miles. 
1  German  short  mile  =3^^^  Eng.  miles. 
1  Eng.  mile  =  lJ+Russian  verst. 

7.  LAND  OR  SQUARE  MEASURE. 

144  Square  inches  make  1   Square  Foot,  marked  In.Ft. 
9  Feet        -        -          1  Yard,         -  -          Yd. 

272*  Feet*  ™  }    "        '     l  Rod'  Pole  °r  Percb'     Rod' 

40  Rods  1  Rood  or  ^  of  an  acre,  Rood. 

4  Roods  -       -     1  Acre,  -         -  Ac. 

640  Acres  1  Mile,         -  -  Mi. 

By  this  measure,  surfaces*  are  measured.      It  is  long 

measure  squared,  or  multiplied  into  itself. 

8.  CUBIC  OR  SOLID  MEASURE. 

1728  Solid  Inches  make  1  Foot,  Marked  In*  Ft. 

27  Feet  1  Yard,  Yd. 

50  Feet  of  hewn,  or  )  t   m  T      j  TT 

40  Feet  of  round  Timber      }  l  Ton  or  Load' 
128  Feet,  i.e.  8  feet  in  length       )     c    d    f  w     d  c 

4  in  breadth  and  4  in  height    ) 

By  this  measure  the  contents  of   solids  are  obtained, 
or  things  that  have  length,  breadth,  and  depth.     It  is  long 
measure  cubed,  or  multiplied  by  itself,  twice, 
C  2 


30                                                 MEASURES. 

9.  DRY 

MEASURE. 

2  Pints  make     -      -    - 

1  Quart,  marked 

4  Quarts 

1  Gallon, 

2  Gallons 

1  Peck,     - 

4  Pecks         - 

1  Bushel,     - 

8  Bushels 

1   Quarter, 

36  Bushels     - 

1  Chaldron,  - 

Pt.  Qt. 

Gal. 

Pk. 
Bus. 

Qr. 

dial 

8  bushels  a  Hogshead  of  Salt. 

NOTE. — The  diameter  of  ihe  Winchester  or  common 
bushel  is  18|  inches,  and  its  depth  8  inches. 

The  gallon  dry  measure  contains  268f  cubic  inches. 

Corn,  grain,  beans,  peas,  flax-seed,  salt,  coals,  &c.  are 
measurecf  by  striked  measure  ;  but  pears,  apples,  turnips, 
potatoes,  onions,  &c.  are  heaped  to  a  handsome  round- 
ing measure.  The  bushel  contains  2150f +  cubic  inches. 

10.  WINE  MEASURE. 

4  Gills  make  -    1  Pint,  marked     -     Gill  Pt. 

2  Pints  1   Quart,  Qt. 

4  Quarts  -       I  Gallon,  -       Gal. 

42  Gallons  -    1  Tierce,     -  Tier. 

63  Gallons         -  1   Hogshead,  -        Hhd. 

84  Gallons     -  1  Puncheon,  Punch. 

2  Hogsheads  1  Pipe  or  Butt,         -        Pipe. 

2  Pipes  or  4  Hhds.     -      1  Tun,  Tun. 

NOTE. — The  wine  gallon  contains  231  cubic  inches. 
The  hogshead  of  63  gallons,  and  the  puncheon  of   84 
gallons,  are  not  used  with  us.    The  hogshead  of  108  or  110 
gallons  is  called  a  hogshead  or  a  puncheon.  Brandies,  spir- 
its, perry,  cider,  vinegar,  mead,  oil  and  honey,  are  sold  by 
this  measure,  though  honey  is|gmetimes  sold  by  the  pound 
avoirdupois.    Milk  is  sometimes  measured  by  this  meas- 
ure, though  more  commonly  and  justly  by  beer  measure. 
11.  ALE  MEASURE. 

2  Pints  make  1  Quart,  marked     -       Pt.  Qt. 

4  Quarts                     -     1  Gallon,  Gal. 

5  Gallons                        1  Firkin  of  Ale,  -        A.  Fir. 

9  Gallons  -                 1  Firkin  of  Beer  -      B.  Fir. 
36  Gallons                         1  Barrel  of  Beer,  Bar. 
54  Gallons  -                    1  Hogshead,  Hhd. 

3  Barrels  1  Butt,  -  Butt. 
NOTE. — The  Ale  gallon  contains  282  cubic  inches. 
Milk  is  sold  by  the  Beer  quart,  which  is  about  one  sixth 

larger  than  the  wine,  cider,  &c.  quart.  32  gal.=  Ibar.  ale* 


TIME,    MOTION.  31 

12.  TIME. 

60  Seconds  make  1  Minute,  marked  8.  M. 

60  Minutes                                     1  Hour,  H. 

24  Hours                                       1  Day,  -       D. 

7  Days                                          1  Week,  W. 

4  Weeks         -             :                1  Month,  -        -      Mo. 

13  Lunar  or  12  Solar  months  )  t    v  v 

nr>r   TV  i    *         JL  CclT,  -  JL  •• 

or  365  Days  J 

NOTE. — 305  days,  5  hours,  48  minutes,  48  seconds, 
make  a  solar  year  according  to  the  most  exact  observation. 

April,  June,  September  and  November,  have  each  30 
days  ;  each  of  the  other  months  has  31,  except  Februaryr 
which  has  28  in  common  years  and  29  in  leap  years.* 

30  years  make  an  age,  and  100  years  a  century, 

A  lunar  month  is  29d.  12h.  44m.  3s.  nearly. 

13.  CIRCULAR  MOTION. 

60  Seconds  make       -        1  Prime  minute,  marked  "  ' 
60  Minutes  -      1  Degree, 

30  Degrees  -       -       1  Sign,  -      & 

12  Signs,  or  )  (  The  whole  circle 

360  Degrees      J  (of  the  Zodic. 

*  To  find  whether  any  given  year  will  be  leap  year. 
RULE. — Divide  the  given  year  by  4  ;  if  nothing-  remain,  it  is 
leap  year ;  but  if  there  be  a  remainder,  that  is  the  number  of 
years  after  leap  year. 

EXAMPLE. 
Was  1823  leap  year  ?  4)1823 

455-3  rem. 
which  shows  it  to  have  been  the  3d  after  leap  year. 

The  last  year  in  every  three  centuries  out  of  four,  which  would 
otherwise  be  leap  year,  is  to  be  reduced  to  a  common  year. 

To  find  whether  the  last  year  in  any  given  century  is  leap  year. 

RULE. — Divide  the  given  century  only,  or  the  hundreds  in  the 
year,  by  4 ;  if  nothing-  remain,  it  is  leap  year ;  but  a  remainder 
shows  it  is  to  be  counted  a  common  year. 

EXAMPLE. 

Will  the  year  1900  be  leap  year  ? 
4)19(4 
16 

3  remainder ;  therefore  a  common  year. 


82  REDUCTION. 


REDUCTION. 


REDUCTION  teaches  to  change  the  denomination  of 
numbers  without  altering  their  value. 

RULE. — Wfcen  the  reduction  is  from  a  higher  denomi- 
nation, to  a  lower,  as  pounds  into  shillings,  tons  into  oun- 
ces, &c.  multiply  the  highest  denomination  by  as  many 
of  the  next  lower  as  make  one  of  the  highest,  adding  to 
the  product  the  parts  of  the  same  name  ;  multiply  this 
sum  by  the  next  lower,  adding  to  the  product  the  parts  of 
its  own  name,  if  any ;  and  so  on  to  the  denomination 
required. 

When  the  reduction  is  from  a  lower  to  a  higher  denomi- 
nation, pence  into  pounds,  minutes  into  days,  &c.  divide 
the  given  number  by  as  many  of  that  denomination  as 
make  one  of  the  next  higher,  and  so  on,  to  the  denomi- 
nation required ;  and  the  last  quotient  with  the  several 
remainders,  (if  any)  will  be  the  answer  required. 

The  proof  is  had  by  reversing  the  question. 


EXAMPLE. 
MONEY. 

1.  In  476  pounds,  how  many  shillings  and  pence  1 
476 
20 

9520  Shillings. 

12  12)114240 

Ans.  114240  Pence.        2,0)952,0 

Proof,  476 

2.  In  3694  shillings,  how  many  pence  ?          Ans.  44328. 

3.  How  many  farthings  in  69217  pence  1     Ans.  276868. 

4.  Reduce  6942  pounds  to  farthings.         Ans.  6664320. 


REDUCTION.  33 

5.  In  c£49  195.  llfd.  how  many  farthings? 

£•   s.  d.  qr. 
49  19  11   3 
20 

999  Shillings, 
12 

11999  Pence. 
4 

Ans.  47999  Farthings. 

6.  How  many  Pence  in  c£472  13s.  4d.  ? 

Ans.  113440, 

7.  How  many  pounds  in  46*216  farthings  ? 

4)467216 

12)116804 
2,0)973,3  Sd. 

Ans.  ,£486  13s.  Sd. 

8.  How  many  pounds  in  9752  pence  1 

Ans.  <£40  12s.  Sd. 

9.  In  648  English  guineas,*  how  many  pence  ?. 

Ans.  217728. 
*  TABLE, 

Showing  the  weight  and  value  of  several  pieces  of  For- 
eign Coins. 

£  s.  d.  $     cts. 

An  English  Shilling  is  -  -  1   4      or  22-f 

A  French  Franc,  -  1    1£  18$ 

A  Livre  Tourriois,         -  .  1   ]J  18J 

An  English  or  French  Crown,        -  6    7.1  110 

Napoleon,         4  dwt.  6  grs.  -  123  3     71 

4  Johannes*     9  -  -  280  8     00 

Moidore,  6  18  1   16    0  6     00 

Eng.  Guinea,  5  6  -     1     8    0  4     662 

French     do.    5  5  -      1     7   3i  4     54£ 

Sp.  Pistole,      4  5         -  113  3     54^ 

By  an  Act  of  Congress  passed  April  29th,  1816,  the  gold  coins 
of  Great  Britain  and  Portugal,  are  estimated  at  27  grains  to  the 
dollar;  those  of  France  at  27^  grains  to  the  dollar;  and  those 
of  Spain  and  her  dominions,  at  28^  grains  to  the  dollar. 


34  REDUCTION. 

10.  How  many  Eng.  guineas  in  28560  pence  ?  Ans.  85. 

11.  In  37  Spanish  pistoles,  how  many  farthings? 

Ans.  37740. 

12.  In  48960  farthings,  how  many  pence,  three-pences, 
six-pences  and  dollars  1     Ans.  12240  pence,  4080  three- 

pences, 2040  six-pences  and  170  dollars. 

13.  In  427  moidores,  how  many  dollars  and  pounds  ? 

Ans.  $2562,  or  ,£768  12s. 

14.  In  11040  pence,  how  many  dollars?  Ans.  $15325. 

15.  How  many  pounds  in  91751  farthings  ? 

Ans.  .£95  11s. 


TROY  WEIGHT. 

1.  How  many  grains  in  47ft.  100z.  of  gold  ? 

Ans.  275520. 

2.  In  47128  grains  of  gold,  how  many  pounds  ? 

Ans.  8ft.  20z.  Sdwt.  16grs. 

3.  In  5605  grains,  how  many  ounces  ? 

Ans.  lloz.  \3dwt.  ISgrs. 

4.  How  many  grains  in  18ft.  5oz.  9dwt.  21grs.  ? 

Ans. 

AyontBtfpots  WEIGHT* 

1.  In  }%cwt  Iqr.  18ft.  how  many  ounces,  at  25ft.  to 
the  qr.  of  a  cwt  ?  Ans.  19888. 

2.  How  many  tons  in  3440640  drams  ?     Ans.  6  Tons. 

3.  In27f.    \5cwt.   2grs.   17ft.   how  many  pounds,  at 
25ft.  to  the  quarter  ?  Ans.  5567ft* 

4.  How  many  pieces  of  4ft.  5jft.  and  6£jfc.  of  each 
an  equal  number,  in  6%cwt.  %qrs.  24ft.  of  beef? 

Ans.  439  pieces  of  each. 

5.  In  3  tons  of  hay,  how  many  pounds  ?  Ans.  6720ft. 

APOTHECARIES'  WEIGHT. 

1.  In  31ft.  2^.  65.  how  many  drams.  Ans.  2998. 

2.  How  many  pounds  in  2535  scruples  1 

Ans.  Sft.  9§.  53. 

CLOTH  MEASURE. 

1.  How  many  quarters  in  83yds.  Sqrs.1  Ans.  335. 

2.  In  2528  nails,  how  many  yards  ?  Ans.  15£. 

3.  In  840  nails,  how  many  ells  English  ?  Ans.  42. 


REDUCTION.  35 

4.  How  many  yards  of  Holland  in  58  pieces,  each  con- 
taining 36  ells  Flemish  1  Ans.  1566. 

5.  In  748  ells  French,  how  many  ells  English,  ells 
Flemish,  yards,  quarters,  and  nails  1         Ans.  897  E.  E. 
3  ^rs.—  1496  E.  Fl— 1122  yds.— 4488  qrs.— 17952  na. 

LONG  MEASURE. 

1.  In  70  miles,  how  many  furlongs  and  poles  ? 

Ans.  560  fur.  22400  poles. 

2.  How  many  leagues  in  21120  yards  7  Ans.  4. 

3.  How  many  barley  corns  in  360  degrees,  each  degree 
69£  miles  ?  Ans.  4755801600. 

4.  How  often  will  a  wheel  that  is  15  feet  in  circumfer- 
ence, turn  round  in  the  distance  from  Hallowell  to  Farm- 
ington,  it  being  32  miles  1  Ans.  11264. 

LAND  OR  SQUARE  MEASURE. 

1.  In  17  acres  3  roods  10  poles,  how  many  poles  ? 

Ans.  2850. 

2.  In  815443200  inches,  how  many  acres  1     Ans.  130. 

3.  How  many  acres  in  6654  rods  or  poles  ? 

Ans.  41  Acres,  2  Roods,  14  Rods. 

4.  In  6  acres  1  rood,  how  many  perches  1  Ans.  1000. 

5.  If  a  room  be  16  feet  long,  and   14  feet  wide,  how 
many  feet  of  boards  will  it  take  to  lay  the  floor  1 

Ans  224  feet. 

6.  How  many  shingles  will  it  take  to  cover  the  roof  of 
a  house  40  feet  in  length,  and  of  18  feet  rafters,  allowing 
each  shingle  to  be  4  inches  wide,  and  each  course  to  be 
laid  out  6  inches  ?  Ans.  8640. 

7.  How  many  boards  will  cover  a  barn  that  is  50  feet 
long,  and  30  feet  wide  ;  the  height  of  the  gable  ends  13 
feet,  and  the  rafters  20  feet  each  ;  and  the  posts,  or  body 
of  the  frame,  15  feet  in  height  ?  Ans.  4790  feet. 

CUBIC  OR  SOLID  MEASURE. 

1.  In  9  tons  of  round  timber,  how  many  inches  ? 

Ans.  622080. 

2.  How  many  cords  of  wood  in  3096576  inches  1 

Ans.  14. 

3.  In  259200  inches  of  hewn  timber,  how  many  tons  ? 

Ans.  3. 

4.  How  many  bricks,  8  inches  long,  4  wide,  and  2 
thick,  will  build  a  house  44  feet  long,  40  feet  wide,  and  20 
feet  high,  with  walls  12  inches  thick  ?         Ans.  88560. 


36 


REDUCTION. 


DRY  MEASURE. 

1.  In  49  bushels,  how  many  quarts  ?  Ans.  1568. 

2.  How  many  bushels  in  -27072  quarts  ?          Ans.  846. 

3.  How  many  pints  in  150  bushels  of.  corn  ?  Ans.  9600. 

4.  In  56    bushels    of    wheat,    Canada   measure,    how 
many  bushels  of  the  United  States  1  Ans.  70. 

NOTE. — 5  pecks,  or  40  quarts,  make  1  Canada  bushel. 

WINE  MEASURE. 

1.  In  3  hogsheads,  how  many  gills  ?  Ans.  6048. 

2.  How  many  hogsheads  in  6480  gills  ? 

Ans.  3  Hhds.  13  Gals.  2   Qts. 

3.  How  many  pints  in  25  t'uns  of  wine  ?     Ans.  50400. 

4.  In  30876  gills,  how  many  hogsheads  1 

Ans.  15  Hhds.  19  Gals.  3  Qts.  1  Pt. 

BEER  MEASURE. 

1.  In  10  hogsheads  17  gallons,  how  many  gills? 

Ans.  17824. 

2.  How  many  firkins  of  ale,  in  7624892  pints  ? 

Ans.  1 19188  A.  Fir.  7  Gals.  2  Qts. 

3.  How  many  pints  in   12  hogsheads,    15   gallons,   2 
quarts  ?  Ans.  5308. 

4.  In  6420  quarts,  how  many  firkins  of  Beer  1 

Ans.  178  B.  Fir.  3  Gals. 

TIME. 

1.  How  many  minutes  in  347  days  1          Ans.  499680. 

2.  In  57953  hours,  how  many  weeks  1 

Ans,  344  W.  6  D.  17  H. 

3.  How  many  seconds  are  there  in  72  years,  10  days, 
18  hours,  11  minutes,  allowing  365  days  and  6  hours  to  a 
year  ?  "  A  ns.  2273076660. 

4.  How  many  days  from  the  20th  of  April  to  the  16th 
of  December  following  ?  Ans.  240. 

5.  Suppose  your  age  to  be  16  years  and  20  days,  how 
many  seconds  old  are  you,  allowing  365  days  and  6  hours 
to  the  year  1  Ans.  506649600  Sec. 

6.  How  many  days  from  the  birth  of  Christ,  to  Christ- 
mas 1823,  allowing  the  year  to  contain  365^  days  ? 

Ans.  665850J  Days. 

7.  In  a  lunar  month,  how  many  seconds  ? 

Ans.  2551443  &«. 


APPLICATION  OF  REDUCTION.  37 

CIRCULAR  MOTION. 

1.  In  6  signs  of   the   zodiac,  through   which  the  sun 
moves  in  half  a  year,  how  many  seconds  ? 

Ans.  648000. 

2.  How  many  prime  minutes  in  360  degrees  1 

Ans.  21600. 

APPLICATION. 

1.  Four  men  brought  each  £70  sterling  value  in  gold 
into  the  mint ;  how  many  guineas  at  21s.  each  must  they 
receive  in  return  1  Ans.  266  guin.  14s. 

2.  A  silversmith  received  3  ingots  of  silver  each  weigh- 
ing 54  ounces,  with  directions  to  make  them  into  spoons 
of  2  oz.,  cups  of  5  oz.,  salts  of  1  oz.,  and  snuff  boxes  of 
2  oz.,  and  deliver  an  equal  number  of  each ;    what  was 
the  number  ?  Ans.  16  of  each,  and  2  oz.  over. 

3.  Suppose  a  ship's  cargo  from  Bourdeaux  to  consist  of 
250  pipes,  130  hhds.  and  150  quarter  casks  or  4  hhds. ; 
how  many  gallons  are  there  in  all ;  and,  allowing  every 
pint  to  be  a  pound,  what  burden  was  the  ship  of? 

Ans.  44415  gal.,  and  the  ship's  burden  ) 
was  158  tons  12  cwt.  2  qrs.  ) 

4.  In  15  pieces  of  cloth,  each  piece  20  yds,  how  many 
French  ells  ?  Ans.  200. 

5.  In  10  bales  of  cloth,  each  bale  12  pieces,  and  each 
piece  25  Flemish  ells,  how  many  yards  ?       Ans.  2250. 

6.  The  forward  wheels  of  a  wagon  are  14^  feet  in  cir- 
cumference, and  the  hind  wheels  15  feet,  9  inches  ;  how 
many  more  times  will  the  forward  wheels  turn  round  than 
the  hind  wheels,  in  running  from  Hallowell  to  Boston,  it 
being  174  miles  ?  Ans.  5028^  times. 

7.  How  many  times  will  a  ship  97  feet,  6  inches  long, 
sail  her  length,  in  the  distance  of  12800  leagues,  and  10 
yards  ?  Ans.  2079508. 

8.  The  sun's  mean  distance  from  the  earth  is  95,000,000 
of  miles  ;  and  a  cannon  ball  at  its  first  discharge,   flies 
about  a  mile  in  7^  seconds  ;  how  long  would   a  cannon 
ball  be,  at  that  rate,  in  flying  from  the  earth  to  the  sun  ? 

Ans.  22yrs.  216days,  12h.  40min. 

9.  If  a  field  be  36  rods  long,  and  24  rods  wide,  how 
many  acres  does  it  contain  ?      Ans.  5  ac.  1  roo.  24  rods. 

D 


68  FEDERAL    MONEY. 

10.  How  many  strokes  does  a  regular  clock  strike  in 
365  days  or  a  year  ?  Ans.  56940. 

11.  How  Jong  will  it  take  to  count  a  million,  at  the  rate 
of  50  a  minute         Ans.  333h.  20m.  or  13d.  2!h.  20m. 

12.  If  the  national  debt  of  England   amounts  to  837 
millions   of  pounds  sterling  ;  how  long  would  it  take  to 
count  this  debt  in  dollars,  (4s.   6d.  sterling,)   reckoning, 
without  intermission,  twelve  hours  a  day,  at  the  rate  of 
50  dollars  a  minute  ;  and  allowing  365  days  to  the  year  1 

Ans.  283  yrs.  38  days.  4  hours. 

13.  In  42  pigs  of  lead,  each  weighing  4cwt.  3qrs.  how 
many  fother,  at  19cwt.  2qr.  ?      Ans.  10  fother,  4^  cwt. 

14.  A  gentleman  has  20  hhds.  of  tobacco,  each  8  cwt. 
3  qrs.  14ft),  and  wishes  to  put   it  into   boxes  containing 
70  Ib.  each  ;  I  demand  the  number  of  boxes  he  must  get, 
at  25ft  to  the  qr.  ?  Ans.  254. 

15.  How  many  coats  can  be  made  out  of  36f  yds.  of 
broadcloth,  allowing  If  yds.  for  a  coat  1  Ans.  21. 

16.  A  man  would  ship  720  bushels  of  corn,  in  barrels 
which  will  hold  3  bus.  3  pks.  each  ;    how  many  barrels 
must  he  get  ?  Ans.  192. 

1  7.  How  many  pints,  quarts,  and  two  quarts,  of  each  an 
equal  number,  may  be  filled  from  a  pipe  of  wine  ? 

*  .  Ans.  144. 

18.  Three  fields  contain,  the  first  7  acres,   the   second 
10  acres,  and  the  third  12  acres,  1  rood;  how  many  shares 
can  they  be  divided  into,  each  share  to  contain  76  perches? 
Ans.  61  shares,  and  44  perches  over. 


FEDERAL  MONEY,* 

THE  denominations  of  Federal  Money,  like  figures  in 
whole  numbers,  increase  in  a  tenfold  proportion,  begki- 
ing  with  mills,  of  which 

10  make       -     1  Cent,  marked  m.  c.  respectively. 

10  Cents      -     1  Dime,     -     -     -     d. 

10  Dimes     -     1  Dollar,  -     -     Doll  or  $. 

10  Dollars         I  Eagle,   -    -    -     E. 

*  Federal  Money  ought,  in  strict  propriety,  to  be  treated  of 
after  decimal  fractions  ;  but  usefulness,  (as  fractions  are  not  al- 


FEDERAL    MONEY.  39 

In  the  money  of  account  the  dollar  is  considered  as  the 
unit :  all  other  denominations  being  valued  according  to 
their  distance  from  the  dollar's  place.  A  point  or  comma 
must  be  placed  after  the  dollars  to  separate  them  from  the 
lower  denominations ;  then  the  first  figure  at  the  right  of 
the  comma  is  dimes,  the  second  cents  and  the  third 
mills ;  but  in  reckoning,  the  two  first  are  called  so  many- 
cents,  using  the  dimes  for  the  tens'  place  of  cents.* 
When  the  cents  in  any  sum  are  less  than  10,  a  cipher 
must  be  put  in  the  place  of  dimes,  or  tens'  place  of  the 
cents,  before  any  operation  is  performed, 

ADDITION  OF  FEDERAL  MONEY. 

RULE. — Place  the  numbers  according  to  their  value, 
dollars  under  dollars,  cents  under  cents,  <fcc.  and  add  as 
in  whole  numbers,  placing  the  comma  in  the  sum  directly 
under  the  commas  above. 

ways  understood)  requires,  and  its  simplicity  and  near  alliance  to  { !\f[ 
whole  numbers,  will  admit  it  in  this  place. 

The  coins  of  the  United  States  are  three  of  gold,  six  of  silver 
and  two  of  copper.  The  gold  coins  are  called  an  eagle,  half 
eagle  and  quarter  eagle ;  the  silver,  a  dollar,half  dollar,  quarter 
dollar,  double  dime,  dime  and  half  dime  ;  and  the  copper,  a  cent 
and  half  cent. 

The  weight  of  the  eagle  is  11  penny  weights  and  6  grains  ;  the 
weight  of  the  dollar  17  pennyweights  8  grains;  of  the  dime,  1 
pennyweight  17  3-5  grains  ;  of  the  cent  8  pennyweights  16  grains. 
The  standard  for  gold  coin  is  eleven  parts  fine  gold  and  one  part 
alloy;  the  alloy  consisting  of  silver  and  copper.  The  standard 
for.  silver  is  1435  to  179  alloy  ;  the  alloy  being  wholly  of  copper. 
Alloy  is  used  in  gold  or  silver  to  harden  it. 

When  either  gold  or  silver  is  finer  or  coarser  than  the  standard, 
the  variation  from  the  standard  is  estimated  by  carats  and  grains 
of  a  carat  in  gold,  and  by  pennyweights  in  silver. 

NOTE. — A  carat  is  not  any  certain  quantity  or  weight^  but^-  of 
any  weight  or  quantity  ;  which  minters  and  goldsmiths  divide 
into  four  equal  parts  called  grCtins  of  a  carat. 

*  Any  sum  of  this  money  may  be  read  differently ;  either  wholly 
in  the  lowest  denomination,  or  partly  in  the  higher  and  partly  in 
the  lowest.  Thus  the  sum  $24,  367  may  be  read  24367  mills  ;  or 
2436  cents.  7  mills  ;  or  243  dimes,  6  cents  and  7  mills  ;  or  24  dol- 
lars, 3  dimes,  6  cents  and  7  mills  ;  or  2  eagles,  4  dollars,  3  dimes, 
6  cents  and  7  mills  ;  or  24  dollars,  36  cents  and  7  mills  ;  but  the 
last  is  the  usual  method. 


FEDERAL    MONEY, 

EXAMPLES. 

I. 

2. 

3. 

c.  m. 
4612,39  4 
5746,29  6 
6154,34  7 
2410,61  8 
3917,46  5 

$    c.  m. 
274,62  1 
125,48  6 
612,59  4 
431,  51  3 
246,34  9 

$  c.  m. 
489,20  6 
217,16  5 
6,74  9 
12,81  6 
7,10 

Sum.        22841,120 

18228,72  6 

Proof.     22841,12  0 

4. 

5. 

6. 

<s> 
$>.     c. 

3792,47^ 
684,16 
59,76 
1246,23} 

$.  c.  m. 
76,28  1 
39,46  2 
57,19  7 
68,49  8 

9.  c. 
180,20 
29, 
6,17 
31,42 

APPLICATION. 

l»  Suppose  that  B  owes  A  $75,  17c  ;  C  owes  15c.  4m  ; 
D  owes  $21,  13c.  6m  ;  E  owes  9c. ;  F  owes  $796,  3c. ; 
and  G  owes  $17,13c. ;  what  is  due  to  A  from  all  of 
them  :  Ans.  $909,71c. 

2.  There  is  a  gallant  ship  just  returned  from  the  Indies, 
which  is  herself  worth  $12l45,86c. ;  and  one  quarter  of 
her  cargo  is  valued  at  $25411,65c. ;  pray  tell  me  the 
value  of  the  ship  and  cargo.  Ans.  $113792,46c. 


SUBTRACTION  OF  FEDERAL  MONEY. 

RULE. — Place  the  less  sum  under  the  greater  in  the 
same  manner  as  in  addition.  Subtract  as  in  whole  num- 
bers, and  place  the  comma  directly  under  those  above. 


FEDERAL    MONEY. 


41 


1. 

$  c.  m. 
From  4612,60  5 
Take  2904,71  6 


EXAMPLES. 
2. 

$     c. 

41,20 

3,97 


3. 

£•  ni. 
317,61  1 
149,72  1 


Rem.  1707,889 


Proof  4612,60  5 


4. 

$  c,  m. 

2910,70  2 

827,06  4 


APPLICATION. 

1.  Suppose  that  my  rent  for  three  months  is  $268,  and 
that  I  have  paid  for  taxes  $58,16c.,  and  for  several  repairs 
$73,85^c. ; — what  have  I  to  pay  of  my  quarter's  rent  1 

Ans.  $135,98*c. 

2.  Jack  Hatchway  received  prize  money  to  the  amount 
of  $1000  ;  he  then  laid  out  $411,410.  for  a  span  of   fine 
horses  ;  $123,40c.  for  a  suit  of  new  clothes,  and    a  gold 
watch  ;  and  $359, 50c.  were  lost  in  lottery  gambling ; — • 
what  will  he  have  left,  after  he  has  paid  his  landlord's  bill, 
which  is  $S5,llc.  ?  Ans.  $20,5Sc. 

MULTIPLICATION  OF  FEDERAL  MONEY. 

.  RULE — Multiply  as  in  whole  numbers,  and  place  the 
comma  as  many  figures  from  the  right  hand  in  the  pro- 
duct as  it  is  in  the  multiplicand. 

EXAMPLES. 

1.  2. 

$ci> 
c.  $     c.  m* 

Multiplicand,  4769,67  4276,96  7 

Multiplier,  36  48 


Product. 


2861802 
1430901 

171708,12 
D2 


205294,41  0 


42  FEDERAL    MONEY. 

$  c.  m.  $    c.  m. 

3.  Multiply  67,48  2     by       5  Product  337,41  0 

4.  76,43  4  305,72 

5.  3,164  9  28,476 

6.  ,78  1  12  9,37  2 

7.  1,06  45  47,70 

8.  3,16  150  474,00 

9.  4,25  598  2541,50 

10.  4,96  3  347  1722,16  1 
NOTE. — To  multiply  by  £,  *,  -| ,  &c.     Take  -J,  \.  |,  &c. 

of  the  multiplicand  first,  and  set  it  down  beneath  the  line 
as  a  product;  then  multiply  by  the  whole  number,  set- 
ting the  product  or  products  below  that  of  the  fraction, 
and  add  all  together. 

11.  10,50  14^  152,25 

12.  1,20  84J  101,10 

13.  2,40  26f  64,20 

APPLICATION. 

1.  What  will  120  yards  of  damask   come  to,  at  $12, 
5c.  per  yard  ?  Ans.  $1446. 

2.  Find  the  amount  of  the  following  bill  of  Parcels. 

Hallowell,  Nov.  1,  1829. 
Mr.  Peter  Paywell, 

Bought  of  Francis  Fairdealer ; 

28  fb>  of  Green  Tea,       at  $2,  15c.  per  ft.         $          c. 
41'  ft.  of  Coffee,  0,  15 

34  ft.  of  Loaf  Sugar,  0,  19 

13  cwt.  of  Malaga  Raisins,    7,  31  per  cwt. 

35  firkins  of  Butter,  7,  14  per  fir. 
27  pairs  of  Worsted  Hose,    1,    4  per  pair, 
94  bushels  of  Oats,                 0,  33  per  bush. 

29  pairs  of  Men's  Shoes,      1,  12  per  pair, 


Amount,— $509,32c. 
Received  payment  in  full, 

Francis  Fairdealer. 

A  SHORT  RULE, 

To  know,  mentally,  the  value  of  100  pounds  of  any  article 
in  Federal  Money,  when  the  price  of  lib.  is  given. 

RULE. — Cnil  the  cents  in  the  price  of  1  pound,  dollars, 
and  that  sum  v/Ul  be  the  value  of  100  pounds  of  the  arti- 


FEDERAL    MONEY.  43 

cle.  If  there  be  several  hundred  pounds  of  it,  multiply 
the  value  of  100ft.  thus  found,  by  the  number  of  hun- 
dreds and  you  will  have  the  answer  accordingly.  The 
whole  may  be  done  by  a  single  glance  of  the  mind. 
Parts  of  a  cent  in  the  price  of  1  pound,  will  be  the  same 
parts  of  a  dollar  in  the  price  of  100  pounds.  100ft.  at 
1  cent  per  ft.=100  cents  =  l  dollar. 

Therefore,  lOOjfr.  of  beef,  at  4  cents  a  ft.  will  come 
to  400  cts. =$4. 

What  will  500ft.  of  pork  come  to  at  8  cts.  a  ft.  ? 

Your  mind  tells  you  8  cts.  a  ft.  is  $8  a  100  pounds  and 
8  multiplied  by  5  is  40  ;  of  course  the  answer  is  $40. 

DIVISION  OF  FEDERAL  MONEY. 

RULE. — Write  the  number  and  divide  as  in  simple  di- 
vision. The  quotient  will  be  of  the  same  denomination 
as  the  lowest  of  the  dividend.  Or,  when  you  have  divid- 
ed the  dollars  of  the  dividend,  put  the  comma  or  point 
in  the  quotient ;  if  the  dollars  be  less  than  the  divisor, 
put  the  point  down  at  first. 

EXAMPLES. 
1.  2.  3. 

$.  c.  m.  $.  c.  $.  c.  m. 

6)47,26  2  8)6914,24  1 1  )7,49  1 


17,87  7 


4. 

-    f .  c.  m.  $.  c.  m. 

237)6742,27  1(28448+ mills,  or  28,44  8  and  95  Rem. 

5.                            6.  7. 

$.  c.  w.                    $.  c.  m.  $.  c.  m. 

387)753,35  7(       359)259,23  7(  475)74,10  0( 

APPLICATION. 

1.  If  131  yards  of  Irish  linen  cost  $49,78c.,  what  is  it 
per  yard  1  Ans.  38c. 

2.  If  140  reams  of  paper  cost  $329,  what  is  that  per 
ream  ?  Ans.  $2,35c. 

3.  If  a  reckoning  of  $25,41c.  be  paid  in  equal  shares 
by  14  persons,  what  do  they  pay  apiece  ?  Ans.  $l,81^c. 

4.  If  a  man's  wages  be  $235,80c.  a  year,  what  is  that 
a  calendar  month  ]  Ans.  $19,65c. 


44  FEDERAL  MONEY. 

5.  The  salary  of  the  President  of  the  United  States,  is 
twenty-five  thousand  dollars  a  year  ;  what  is  that  a  day  1 

Ans.  $68,49c.+ 

6.  If  the  amount  of  the  Public  Debt  of  the  United 
States,  be  $91,680,090  ;  how  much  would  that  be  for  each 
person  to  pay,  allowing  the  number  of  inhabitants  in  the 
United  States  to  be  eleven  millions?  Ans.  88,33c.  4m.  4- 

7.  Divide  $57  into  120  equal  shares.  Ans.  47Jcents. 


A  SHORT  RULE. 

When  the  value  of  lOOjfo  of  any  article  is  given,  to 
find,  mentally,  what  it  is  per  ffo. 

RULE,  —  Call  the  dollars  in  the  price  of  lOOjfj.  cents, 
and  that  is  the  answer.  If  the  value  of  several  100}^.  is 
given,  first  divide,  in  your  mind,  the  dollars  in  the  price 
of  the  hundreds,  by  the  number  of  hundreds,  and  the 
quotient  will  be  the  price  of  one  hundred,  in  dollars, 
which  call  cents,  as  before  directed,  and  that  will  be  the 
answer.  When  the  price  of  the  100ft,.  in  either  case, 
contains  parts  of  a  dollar,  the  parts  of  a  dollar  become 
parts  of  a  cent  in  the  price  of  the  ]fo. 

100ft)  valued  at  85=100^5  valued  at  500  cents  ;  and 
500  cents-r-  100=5  cents.  Therefore  if  lOOffc.  of  beef 
cost  6  dollars,  or  600  cents,  it  cost  6  cts.  a  ffe. 

What  will  a  ffo  of  pork  be  worth,  when  SOOJfo  are 
worth  $48. 

Your  mind  tells  you  800fc  for  $48  will  be  lOOffe  for 
$6,  or  $48  divided  by  8,  the  number  of  hundreds,  will 
give  $6  for  the  quotient,  or  price  of  1  hundred^;  of  course 
the  price  of  a  pound,  or  the  answer,  is  6  cents. 

REDUCTION  OF  FEDERAL  MONEY. 

To  reduce  Dollars  to  Cents  and  Mills. 
Multiply  the  dollars  by  100  for  cents,  and  the  cents  by 
10  for  mills  or  to  the  dollars  annex  two  ciphers  for  cents 
and  three  for  mills. 

To  reduce  Mitts  and  Cents  to  Dollars. 
Divide  the  mills  by  10,  and  the  quotient  will  be  cents  ; 
divide  the  cents  by  100,  and  the  quotient  will  be  dollars  ; 


FEDERAL    MONEY.  45 

or  if  the  number  be  cents,  point  off  two  ;  and  if  mills, 
three  figures  on  the  right  hand;  then  the  figures  on  the 
left  hand  of  the  comma  will  be  dollars,  the  two  first  on 
the  right  hand  will  be  cents,  and  the  third,  if  any,  will 
be  mills. 

EXAMPLES. 

1;  Reduce  674  dollars  to  cents  and  mills. 
674 
100 


67400  Cents 
10 

674000  Mills. 


Or     67400  Cents.  )  A 
674000  Mills.     J  * 

2.  How  many  dollars  in  642179  mills  ? 
10)642179 


100)64217-9 

$642-17-9    Or   $642  17c.  9m.  Ans. 

3.  How  many  mills  in  47692  dollars  ?   Ans.  47692000. 

4.  In  46791  cents,  how  many  dollars  ?  Ans.  $467,91c. 

5.  In  6421796  mills,  how  many  dollars,  cents  and  mills  ? 

Ans.  $6421  79c.  6m. 

To  reduce  New-England  currency  to  federal  Money. 
CASE  1. — If  the  sum  consist  of   pounds  only,  annex 
three  ciphers  to  it  and  divide  by  3  ;  the  quotient  will  be 
the  answer  in  cents.* 

EXAMPLES. 

1.  Reduce  £3762  to  Federal  Money. 
3)3762000 


1254000  Cents,  or  $12540  Ans. 

2.  Reduce  ,£471  to  Federal  Money.  Ans.  $1570. 

3.  Reduce  .£37  to  dollars  and  cents.     Ans.  $123  33±c. 

*  As  a  dollar  is  2iy  orf^-  of  a  pound,  it  is  plain  that  annexing 
a  cipher  to  the  pounds,  and  dividing-  by  3,  will  give  a  quotient  in 
dollars ;  and  annexing-  other  ciphers,  and  dividing-  by  3  will  give 
tenths,  hundredths,  &c.  of  a  dollar ;  or  dimes,  cents,  &c. 


46  FEDERAL    MONEY. 

CASE  II. — If  pounds  and  shillings  are  given,  to  the 
pounds  annex  half  the  number  of  shillings,  and  two  ci- 
phers, if  the  number  of  shillings  be  even  ;  but  if  the 
number  be  odd,  annex  half  the  even  number,  and  then  5 
for  the  odd  shilling,  and  one  cipher,  and  divide  by  3 ;  the 
quotient  is  the  answer  in  cents. 

EXAMPLES. 

1.  Reduce  £64  16s.  to  dollars  and  cents. 

3)64800 

Ans.  21600c£s.  or  $216. 

2.  How  many  dollars  in  ,£41  14s.  ?  Ans.  $139. 

3.  In  <£26  Is.  how  many  dollars  and  cents  ? 

Ans.  g86,83£c*s. 

4.  How  many  dollars  in  £1  17s.?      Ans.  $6,16fc*s. 

CASE  III. — If  there  are  shillings,  pence,  &c.  in  the 
given  sum,  annex  for  the  shillings  as  before,  and  to  these 
add  the  farthings  contained  in  the  pence  and  farthings  ; 
observing  to  increase  their  number  by  1  when  they  exceed 
12,  and  by  2  when  they  exceed  36 ;  and  divide  as  before. 

EXAMPLES. 

1.  Reduce  c£34  165.  4Jrf.  to  Federal  Money. 

3)34819 

Ans.  11606^5.  or  $116,06^5. 

2.  In  c£200l  Is.  3±d.  how  many  dollars  ? 

Ans.  $6670,21fc*s. 

3.  In  c£591  11s.  9%d.  how  many  dollars  ? 

Ans.  $1971,96fc*s. 

To  reduce  FEDERAL  MONEY  to  New-England  currency. 

CASE  I. — When  the  sum  is  dollars  only,  multiply  by  3  : 
and  double  the  product  of  the  first  figure  for  shillings, 
and  the  rest  of  the  product  will  be  pounds. 

EXAMPLES. 

1.  Reduce  473  dollars  to  New-England  currency. 
473 
3 


Ans.  c£141  18s. 


FEDERAL    MONEY.  47 

2.  How  many  pounds,  &c.  in  579  dollars? 

Ans.  ,£173  14s. 

CASE  II. — When  there  are  cents  in  the  given  sum, 
multiply  the  whole  by  3,  and  cut  off  three  figures  of  the 
product  to  the  right  hand  as  a  remainder  ;  multiply  the 
remainder  by  20,  and  cut  off  as  before :  proceed  in  the 
same  manner  through  the  several  parts  of  a  pound,  and 
the  numbers  standing  on  the  left  hand  make  the  answer 
in  the  several  denominations. 

NOTE. — If  there  be  mills,  cut  off  four  figures,  and 
proceed  as  before. 


EXAMPLES. 

1.  Reduce  $376,27c£s.  to  New-England  currency. 
376,27 
3 


.£112,881 
20 

s.  17,620 
12 


d.  7,440 
4 

qr.  1,760 
Ans.  <£112     17s. 

2.  Reduce  $609,88£cfc>.  to  New-England  currency. 

Ans.  <£182  19s. 


3.  How  many  pounds,  shillings,  &c.  in  $429,21cfc.    5 
mills  1 

429,215 
3 


,£128,7645,  <fcc.     Ans.  ,£128  15s. 


48  COMPOUND    ADDITION. 

PRACTICAL  QUESTIONS  IN  FEDERAL  MONEY. 

1.  Having  borrowed  one  hundred  dollars,  arid  paid  at 
one  time  seventy  dollars,  and  at  another  time  sixteen  dol- 
lars, seven  cents  ;  what  is  still  due  ?      Ans.  $13,93c?s. 

2.  What  will  25  bushels  of  corn  come  to,  at  92  cents 
per  bushel  1  Ans.  $23. 

3.  What  will  376  pounds  of  butter  come  to,  at  18 
cents  per  pound  ?  Ans.  $67,68c£s. 

4.  What  will  39  bushels  of  wheat  come  to,  at  1  dollar 
75  cents  per  bushel  ?  Ans.  $68,25c£s. 

5.  If  Iron  cost  6  dollars  50  cents  per  cwt.,  what  is  it 
per  pound,  25jfc.  to  the  qr.  ?  Ans.  6±cts. 

6.  If  240  pounds  of  pork  come  to  28  dollars  80  cents, 
what  is  it  per  pound  ?  Ans.  12c;te. 

7.  Borrowed  607  dollars  20  cents  ;  paid  ,£127  16s.  9d.  ; 
what  is  the  balance  ?  Ans.  $181,7£c£s. 

8.  Lent  400  dollars  ;  received  150  bushels  of  wheat  at 
10s.  per  bushel,  and  200  pounds  of  butter   at  17^  cents 
per  pound,  in  payment  :  how  much  is  still  due  ? 

Ans.  $115. 

9.  What  will  55  yards  of  linen  come  to,  in  Federal 
Money,  at  3s.  9d.  per  yard,  New-England  currency  1 

Ans.  $34,37rts.  5mitts. 

10.  What  will  36  yards  of  broadcloth  come  to,  in  New- 
England  currency,  at  6  dollars  25  cents  per  yard  ? 

Ans.  <£67  105. 

11.  What  will  7  tons  of  hay  come  to,  at  16  dollars  75 
cents  per  ton  ?  Ans.  $117  25rts. 

12.  What  is  the  price  of  one  cwt.  of  hay  at  16  dollars 
75  cents  per  ton  ?  Ans.  83cts. 


COMPOUND  ADDITION. 

COMPOUND  ADDITION  in  Arithmetic,  teaches  to  collect 
several  numbers  of  different  denominations  into  one  sum  ; 
as,  pounds,  shillings  and  pence,  &c.  ;  tons,  hundreds, 
and  quarters,  &c. 

RULE.  —  Place  the  numbers,  so  that  those  of  the  same 
denomination  may  stand  directly  under  each  other,  ac- 


COMPOUND    ADDITION.  49 

cording  to  what  you  were  told  in  Simple  Addition.  Add 
the  first  column  or  lowest  denomination  together,  as  in 
Simple  Addition,  also.  Find  how  many  units  of  the  next 
higher  denomination  are  contained  in  the  sum,  by  divid- 
ing it  by  so  many  of  this  name  as  make  one  in  the  next 
greater.  Set  down  the  remainder,  or  overplus,  under  the 
column  added,  and  carry  the  ones  or  units  (the  quotient) 
to  the  next  denomination,  whose  sum  you  must  find,  and 
proceed  with  as  before  ;  and  so  on  through  all  the  denom- 
inations to  the  highest,  which  add  as  in  Simple  Addition 
setting  down  its  whole  sum.* 

EXAMPLES. 

MONEY. 

£.  s.  d.  £.  s.  d.  qr.  £.  s,  d. 

4210  1611£  97  12  6  2  63  9  4J 

1729  17  9  19  14  5  3  8112  7 

56  11  34-  46  17  9  1  27  16 

207  18  6f  221910  2  1614 

14  6  0£  57  1  2  0  41  9  5 


Sum.  6219  10    6f 
2008  13    7! 
Proof621910    6 


APPLICATION. 

1.  Bought  a  quantity  of  goods  for  c£125  10s. ;  paid  for 
truckage  forty-five  shillings,  for  freight  seventy-nine  shil- 
lings and  sixpence,  for  duties  thirty-five  shillings  and  ten 
pence ;  and  my  expenses  were  fifty-three  shillings  and  nine 
pence  ;  what  did  the  goods  stand  me  in  1  Ans.  <£136  4s.  \d. 

*  The  reason  of  this  rule  is  evident ;  for,  in  addition  of  money, 
as  1  in  the  pence,  is  equal  to  4  in  the  farthings — 1  in  the  shil- 
lings to  12  in  the  pence — and  1  in  the  pounds,  to  20  in  the  shil- 
ling's, to  carry  as  directed  is,  therefore,  only  to  arrange  properly 
the  money,  arising  from  each  column,  in  the  scale  of  denomina- 
tions :  and  this  reasoning-  will  hold  just  in  the  addition  of  com- 
pound numbers  of  any  kind  whatever. 
E 


50  COMPOUND    ADDITION. 

2.  A  prize  being  sold,  and  the  sum  divided  equally 
among  the  captors,  who  were  6  in  number,  each  man  re- 
ceived two  hundred  and  forty  pounds,  thirteen  shillings 
and  seven  pence  ;  what  did  the  prize  cost  the  purchaser  ? 

Ans.  ,£1444  Is.  Qd. 

TROY  WEIGHT. 

jfe.  oz.  dwt.  gr.  ft.  oz.  dwt.  gr. 

671  10  16  13  47      6  11     17 

392    8    9  21  56    11  15    19 

249    7  12  12  63      9    8    22 

516    4    3     7  26      5  19    16 

627    5  17  16  21      6  12    14 


APPLICATION. 

A  gentleman  bought  of  a  silversmith,  dishes  to  the 
weight  of  23ft  6oz.  5dwt. ;  plates  41jfo.  7oz.  17dwt. ; 
spoons  12|fc.  15dwt, ;  salts  2ft.  7oz. ;  waiters  13ft. ;  and 
tankards  7ft.  17dwt. ; — what  weight  of  plate  did  he  buy 
in  all  1  Ans.  99ft.  lOoz.  14dwt. 

AVOIRDUPOIS  WEIGHT. 

Ton.  cwt.  <?r.ft.  oz.  dr.  Ton.  cwt.  qr.  ft 

371   12  1    20  10  13  42  19    3    24 

123  14  2    15     6  7  89  10    2    13 

407     9  3    12  11  9  16     8    1    17 

513  13  0    26    9  15  27  14    0    22 

624  15  1    17     6  11  50     3   2    15 


Here  25ft.  a  qr. 


APPLICATION. 

1.  A  merchant  buys  four  bags  of  hops,  of  which  No.  1, 
weighs  2cwt.  2qr.  10ft.     No.  2,  2cwt.  Iqr.  16ft.    No.  3, 
2cwt.  24ft.    No.  4.  Iqr.  16ft.     He  buys  also  a  couple  of 
pockets  of  hops,  which  weigh  58Jft.  each.     How  many 
hundred  weight  has  he  to  pay  carnage  for,  on  bringing 
them  to  town  ?  Ans.  8cwt.  2qr.  15ft. 

2.  A  country  shopkeeper  buys  of  a  merchant  in  Hallo- 
well,  teas  weighing  3qrs.  14ft. ;  coffee,  Iqr.  23ft. ;  sugars, 


COMPOUND    ADDITION.  51 

3ewt.  2qr.  5ft. ;  spices  2qr.  3ft.  13oz. ;  hops  13cwt.  Iqr. 
24ft. ;  and  several  other  things  to  the  weight  of  3cwt.  17ft 
7oz. ;  for  what  Weight  has  he  to  pay  carnage  on  bringing 
them  home, — 25ft.  a  qr.  ?  Ans.  22cwt.  12ft.  4oz. 

APOTHECARIES'  WEIGHT, 

ft-  §•  3-  9-^r.  ft-  I-  3-  9-s*-. 

26  10  7  2  13  17  9  4  1  14 

54  7  2  1  12  55  10  6  2  10 

76  8  3  6  14  61  4  2  1  9 

83  9  4  2  6  92  11  5  0  18 

41  6  0  1  19  21  6  3  1  17 


APPLICATION. 

An  apothecary  made  a  composition  of  5  ingredients, 
the  1st  of  which  weighed  13ft.  7oz.  ;  the  2d,  lloz.  7dr. 
ISgr. ;  the  3d,  7ft.  2scr. ;  the  4th,  lift.  3dr.  Iscr. ;  and 
the  5th  weighed  15ft.  5oz.  7gr. ; — what  was  the  weight 
of  the  whole  1  Ans.  48ft.  3dr.  Iscr. 

CLOTH  MEASURE. 

Yd.  qr.  na.  E.E.  qr.  na.  E.Fr.  qr.  na.  E.FL  qr.na. 

46    1    2  '  74    2    3  86      5    2        29     1  2 

12    3    3  51    4     1  61      4     1         17     2  3 

62    1    1  24    1     2  52      3    9        20     1  2 

83    2    0  56    3     1  24     2     1         75     0  1 

41    0    1  31    1     2  10      0     3        46     1  3 


APPLICATION. 

1.  Having  bought  four  parcels  of  cloth,   the  first  of 
which  contains  25yds.  3qrs.  ;  the  2d,  37yds.  2qrs.  3na. ; 
the  3d,  14yds.  Ina.  ;  and  the  4th,  23yds. ;  I  desire  to  be 
informed  by  you  how  many  yards  are  in  them  all  ? 

Ans.  100yds.  2qrs. 

2.  I  have  six  parcels  of  cloth  ;  the  first  contains  3  E. 
Fl.  and  3na. ;    the  2d,  14yds,  Iqr.  ;  the  3d,  15E.  E.  2qr. 
2na.  the  4th,  Iqr.  3na. ;  the  5th,  19E.  Fr. ;  and  the  6th, 
255yds. ; — how  many  yards  are  there  in  all  1 

Ans.  320yds. 


52  COMPOUND  ADDITION. 

LONG  MEASURE. 

Deg.  mi.fur.pol  ft.  in.  bar.    Mi.  fur.  rod.  yd.  ft. 

2i  17  2  26  12  10  2     62  1  36  4£  2 

46  58  6  19  14  6  1     75  3  24  3  1 

18  62  4  31  10  7  0     49  5  12  2£  2 

21  19  7  14   8  8  2     14  4  17  1  0 

62  37  1   29   7  *6  1     25  2  10  3  2 


APPLICATION. 

1.  From  A  to  B  is  3mi.  2fur.  7pol. ;  from  B  to  C  is  17 
mi.  13pol. ;  from  C  to  D  is  7fur.  and  from  D  to  E  is  5 
mi.  33pol. ; — what  is  the  distance  between  A  and  E  1 

Ans.  26mi.  2fur.  13pol. 

2.  A  person  rode  four  days  ;  on  the  1st  day  he  went  39 
mi.  6fur. ;  the  2d  day,  46mi.  24pol. ;  the  3d,  60mi.  4fur. 
39pol. ;  and  the  4th  day  he  went  but  37mi.  6fur. ;  what 
was  the  whole  distance  of  his  journey  ? 

Ans.  I84mi.  Ifur.  23pol. 

LAND  OR  SQUARE  MEASURE. 

Acre.  rood.  rod.  yd.  ft.  in.  Acrc.rood.  rod. 

271     1      27    16     4  110  21     2      37 

424    2      3i    21  £   2  96  52     1      24 

512    3      16    10     7  71  28     1      16 

327    2      21      9      1  120  43     3     31 

146     1       12  21      6  74  65    0     20 


APPLICATION. 

1.  There  are  5  lots  of  land,  the  1st  of  which  measures 
I3ac.  3roo.  I4rods;  the  2d,  27ac.  29rods ;  the  3d,  I9ac. 
Iroo. ;  the  4th,  3roo.  34rods  ;  and  the  5th,  45ac.  2roods, 
1 1  rods  ;  what  ground  do  they  all  contain  ? 

Ans.  I06ac.  3roods,  8rods. 

2.  A  gentleman  dividing  his  estate  among  4  sons,  gave 
the  1st,  I50acres,  1  rood,  39rods  ;  the  2d,  100acres,25rods; 
the  3d,  75acres  :  and  the  4th,  55acres,  Irood,  I6rods; — 
how  large  was  the  whole  farm  ?  Ans.  381acres. 


COMPOUND  ADDITION.  53 
CUBIC  OR  SOLID  MEASURE. 

T.round.  ft.  inch.      T.liewn.  ft.  Cord.  ft. 

21   36   476        67    30  36  102 

62   17   965        24    27  6i  90 

47   19   8.16        62    14  52  76 

31   28  1146       39    36  9  120 

19   34  1452        17    20  12  34 


AP  PLICA  TION. 

1.  Required  the  solid  contents  of  four  pieces  of  round 
timber,  the  first  of  which  measured  2tons,  39ft.  94in. ;  the 
2d,  4tons,  1695in. ;  the  3d,  35ft.    1183in. ;    and  the  4th, 
Iton,  25ft.  Ans.  9T.  20ft.   1244in. 

2.  Bought  of  A  14cor.    1727sol.  in.    of   wood  ;    of  B 
19cor.  127sol.  ft, ;  of  C  7eor.  98ft.  llOlin. ;  of  D  9cor. ; 
and  of  E  63ft.  1210in.  ;  how  much  did  I  buy  in  all  ? 

Ans.  51c.  34ft.  5S2in.  or  51c.  2|ft.+Wood  Measure, 

DRY  MEASURE. 

Chal.bus.plf.  qt.pt.  Bus.pk.  gal.  qt.pt, 

39     4      1     6     I  62     2      1     3     1 

57     6      3     4     0  79     3      0     2     1 

61     7231  42     2      1     1     1 

52     4      2     5     1  17     1       1     2     0 

42     1      1     2     1  86     0      1     0     1 


APPLICATION. 

1.  A  corn-merchant  sends  over  the  sea,  of  wheat  130 
bus.  3pks.  Igal.  Ipt. ;  of  oats  290bus.  ]gal.  3qts.  ;  of  rye 
he  has  sent  300bus.  3qts.  ;  of  pease  80bus.  3pks. ;  and  of 
beans  50bus. ;  for  what  number  of  bushels  does  he   pay 
freight  1  Ans.  851  bus.  3pk.  Igal.  2qt.  Ipt. 

2.  How  many  chaldrons  of  coals  are  there  in  four 
Joads,  the  1st  containing  l4chal.27bus.2pk.  Iqt. ;  the  2d, 
9chal.  3pk. ;  the  3d,  29chal.  3qts. ;  arid  the  4th,   35chal. 
7bus.  2pk.  4qts.  ?  Ans.  88chal. 


54  COMPOUND  ADDITION. 

WJNE  MEASURE. 

Tun.hhd.gal  qt.  pt.gilL  Jfhd.  gal.  qt. 

2  .  1     42     1     0    3  68  31  3 

5  3    51     2     1     1                47  59  1 
7     2    60     1     1     2                29  48  1 
9     1.   26     1     0    3                68  37  0 

6  1     17    3     1     1  47  18  1 


APPLICATION. 

1.  A  gentleman  bought  oC  a  wine-merchant,   of  port 
wine,  1  tun,  3hhds. ;  of  claret,  3hhds.  47gal. ;  of   moun- 
tain, Ihhd.  5gal.  ;  and  of  Lisbon,  2hhds.  23gal. :    what 
quantity  did  he  buy  in  all  1     Ans.  3tuns,  2hhds.   I2gal. 

2.  Imported  from  Cadiz,  Lisbon,  Oporto,  and  Madeira, 
4  lots  of  wine,  the  1st  of  which  contained  3tuns,   3hhds, 
61gal.  2qt.   Ipt. ;    the  2d,  2pipes,    Ihhd.   48gal.  3gills ; 
the  3d,  2hhds.  41  gal.  1  gill;  and  the  4th,  6pipes,  3hhds. 
38gal.  Iqt. :  how  much  did  I  have  in  all  ?  Ans.  10  tuns. 

ALE  OR  BEER  MEASURE. 

Hhd.gaLqt.  A.fr.galqt.  Butt.bblgalqt. 

416    29    3  69     7     2  25     1     24    1 

34    17    2  43    5     1  36    2     30    2 

178    15     1  53    4     I  81     0     16    0 

315    47     1  46    3     3  74     1     25    1 

351    34     1  29     1     1  98     1     18    1 


APPLICATION. 

1.  A  beer-brewer  has  sent  into  the  country,  ale,  as  fol- 
lows, viz.  at  one  time  3hhds.  I4gal.  3qt. :    at   another,  2 
hhds.  I6gal. ;  at  another,  I4hhds.  27gals.  2qts/;    and  at 
another,  6hhds.  3qts.  :  how  much  was  sent  in  all  ? 

Ans.  26hhds.  5gal. 

2.  Bought  4  lots  of  beer;,*he  1st  contained  3  butts,    I 
bbL3qts. ;  the  2d,    Ibutt,  34gaJ.    Iqt.;  the   3d,  2bbls.  2 
qts. ;  and  the  4th,  5butts,  Ibbl.  2qts%;  what  did  I  buy  in 
all?  Ans.  lObutts,  2bbls. 


COMPOUND    ADDITION.  55 


Fr. 
37 

46 

58 
61 
94 

m. 
10 
9 
7 
5 
12 

w. 
2 
3 
1 
0 
3 

a. 

6 
2 
4 
5 
1 

A.  TW. 
1731 
1447 
13  19 
1250 
21  39 

TIME. 
s. 
37 
25 
31 
47 
56 

Yr.  days, 
41  276 
81  310 
47  163 
2t  360 
72  176 

h. 
20 
17 
8 
19 
14 

m. 
30 
48 
29 
50 
33 

APPLICATION. 

1 .  Peter's  father  was  28  years  old,  (reckoning  1 3  months 
to  1  year,  and  28  days  to  1  month,)  when  Peter  was  born  ; 
betwixt  Peter  and  John  were  2yrs.   10m.    I6d. ;  between 
John  and  James,  one  year  and  eleven  months ;  and  between 
James  and  Job,  3yrs.  7m.  25d.  ;  when  Job  is  I6yrs.  9m. 
27d.  old,  how  old  is  their  father  1    Ans.  53yrs.  I2days. 

2.  Four  memorable  events  occurred,  between  the   1st 
and  2d  of  which  were  2lyrs.   236d.   7h.  40m.  50s.  ;    be- 
twixt the  2d  and  3d,  48yrs.  l9h.  53s. ;  and  between  the  3d 
and  last,  30yrs.  128d.  3h.  18m.  17s.  ;  now,  allowing  365d. 
and  6  hours  to  the  year,  I  wish  to  know  what  'period  of 
time  elapsed  between  the  first  and  last  of  those  events  1 

Ans.  100  years. 
CIRCULAR  MOTION. 

g      O  t         '  H  9  f  If 

2  15  42  57  41  54  39 
5  19  55  33  64  27  21 
1  22  47  28  52  36  42 

3  10  20  12  78  19  16 
1  9  11  24  93  25  34 


APPLICATION. 

1.  If  the  sun's  motion  in  the  zodiac  be  one  day,   1°   1' 
10";  on  another,  the  same;  on  the   next,    1°  1'  11";  on 
the  next,  1°  1'  10"  ;  and  on  the  next  two,  1°   1'  11"  each 
day ;  what  distance  does  he  move  in  the  six  days  1 

Ans.  6°  7'  3". 

2.  If  the  moon's  motion  in  the  signs  be,  on  one  day,  13° 
28'  4";  on   another,    13°   2'   23";    on    another,   12°   40' 
1";  on  another,  12°  2i>  47":    on   another,    12°   ?'  50" ; 
and  on  another,    11°   56'   7";    what   distance   does   she 
move  in  the  six  days  1  <  Ans.  2&  15°  38'  12*, 


66  COMPOUND    SUBTRACTION. 

COMPOUND  SUBTRACTION. 

COMPOUND  SUBTRACTION  is  finding  the  difference  be- 
tween two  numbers,  of  which  one  or  both  are  compound. 

RULE. — Set  the  less  number  under  the  greater,  as  di- 
rected in  Compound  Addition.  Then,  beginning  at  the 
least  denomination,  subtract  the  under  number  of  each 
from  the  upper,  writing  their  respective  remainders  below 
them.  But  if  the  under  number  of  any  of  the  denomi- 
nations be  greater  than  the  upper,  add  so  many  to  the 
upper  as  make  one  of  the  next  higher  denomination  ;  then 
take  the  under  number  from  that  sum,  writing  down  the 
remainder  as  before,  and  carry  or  add  one  to  the  under 
number  of  the  next  higher  denomination  before  you  sub- 
tract it. — The  method  of  proof  is  the  same  as  in  Simple 
Subtraction. 

EXAMPLES. 

MONEY. 

£.    s.     d.  £.    s.      d.  qr.  «£.  5.    d. 

From  691   12   6^  34  11  4    1  81  17  6£ 

Take  284  15    9f  17  14  10  3  21  12  4£ 

Rem.  406  16    82 


Proof.  691  12    6£ 

AP  PLICA  TION. 

1.  What  sum  added  to  .£17  11s.  8jd.  will  make  ,£100  ? 

Ans.  ,£82  8s.  3d.  3qrs. 

2.  Borrowed  .£50  10s.  ;  paid  again,  at  one  time,  ,£17 
11s.  4d.  ;  at  another,  <£9  4s.  8d.  ;  at  another,  £7  9s.  6d.  ; 
and  at  another,  19s.  6^d.  ;  how  much  remains  unpaid  ? 

Ans.  £15  4s. 


TROY  WEIGHT. 

jfc.    oz.  dwt.  gr.  ft.  oz.  dwt.  gr. 

39      8     14     16  71    9  16     11 

16    10     10     19  35     1  17    20 


APPLICATION. 

Sent  to  a  Silversmith  9ffe.  8oz.   14gr.  of  silver  to  be 
wrought ;  he  makes  me  four  dozen  of  spoons,    weighing 
8oz.  6dwt.  21gr. ;  how  much  silver  is  left  1 

Ans.  lloz.  13dwt  17gr. 


COMPOUND  SUBTRACTION.  57 

AVOIRDUPOIS  WEIGHT. 
Ton.  cwt.  qr.  lb.  Cwt.  qr.  lb.  oz.dr. 

73     11     1     21  94     2     11    8    9 

39     17    2      12  47     3    17  11  10 


Here  251b.  a  qr. 


APPLICATION. 

1.  Bought  17cwt.  2qr.  141b.  of  sugar,  of  which  I  sold 
9cwt.  3qr.  251b. ;  how  much  remains  unsold  1 

Ans.  7cwt.  2qr.  171b. 

2.  Bought,  at  one  time,  9cwt.  3qrs.  211b.  8oz.  of  iron, 
and  sold,  next  day,  8cwt.   Iqr.  241b.   14oz. ;    bought  at 
another  time,  15cwt.  131b.  15oz.  of  the  same  kind,  and  sold 
the  same  week,  15cwt.  Iqr. ;  what  remains  unsold  of  the 
two  parcels,  251b.  a  qr.  ?         Ans.  Icwt.  Iqr.  lOlb.  9oz. 

APOTHECARIES'  WEIGHT. 
lb.     1.   5-  9-  gr.  lb.     f.  3-  9-  gr. 

81     7    2    0    11  69     10  4    1     12 

47    2    4    2    15  30     11  1    2     17 


APPLICATION. 

Bought  of  an  apothecary  sundry  articles,  weighing 
18ft.  5^  2Q  ;  he  compounded  two  parcels  from  them, 
one  of  which  weighed  7ft.  75.  19gr.,  and  the  other,  4ft. 
45.  l^.  Igr. :  how  much  was  left  uncompounded? 

Ans.  2ft.  15. 

CLOTH  MEASURE. 

Yd.  qr.  na.    E.  FL  qr.  na.    E.  E.  qr.  na.    E.  Fr.  qr.  na. 
42  1    2  74  1     3  21  3    1  89    4  2 

17  2    1  40  2     1  942  12    0  0 


APPLICATION. 

1.  From  a  fashionable  piece  of  cloth,  which  contained 
52yds.  2na.,  a  tailor  was  ordered  to  take  3  suits,  each 
6yds.  2qrs. ;  how  much  remains  of  the  piece  ? 

Ans.  32yds.  2qrs.  2na. 


58  COMPOUND    SUBTRACTION. 

2.  Swapped  with  John  Jones  a  piece  of  Irish  linen 
containing  36yds.  Iqr.  3na.  for  a  piece  of  Holland  con- 
taining 20E.  Fl.  2qrs.  Ina.  :  and  swapped  also  with  Seth 
Sears  a  piece  of  sheeting  measuring  41E.  E.  4qrs.  2na. 
for  a  piece  of  French  cambric  measuring  30E.  Fr.  5qs. 
3na. ;  each  agreed  to  pay  me  the  balance,  in  quantity,  in 
jean  :  how  many  yards  of  jean  must  I  receive  from  both  ? 

Ans.  26yds.  3qrs.  Ina. 

LONG  MEASURE. 

Deg.  mi.  fur.  pol.ft.  in.  bar.         Mi.fur.poL  yd.  ft* 
81  30    4     24  12   6     1  74  3     16    4     1 

29  41    5     13  14   12  48  5    37    2    2 


APPLICATION. 

1.  Paul  travelled  three  days,  going  33mi.  5fur.  37pol. 
2yd.  2ft.  each  day  ;  Amos  set  out  with  him,  but  travelled 
only  28mi.  7fur.  lyd.  2ft.  each   day  ;  at  the  end  of  the 
third  day,  how  far  was  Amos  behind  Paul  ? 

Ans.  14mi.  4fur.  31pol.  3yd. 

2.  The  ship  Sea-horse,  bound  to  a  port  at  320  leagues' 
distance,  sailed  6  days  at  an  average  rate  of  120mi.  18pol. 
9in.  every  24  hours  ;  at  the  end  of  the  six  days,  how  far 
short  of  her  destined  port  was  she  1 

Ans.  791eag.  2mi.  5fur.  llpol.  12ft. 

LAND  OR  SQUARE  MEASURE. 

Acre.  roo.  per.  yd.  ft.  in.  Ac.  roo.  per. 

93      2     27     14     7  101  39     2     17 

64      3     14     16^  8     97  16     3     19 


APPLICATION. 

1.  Rufus,  David,  Moses,  and  Robert  owned,  together, 
542ac.  Iroo.  34rods,  29^yds.  7ft.  142in.  of  land ;  the 
first  had  lOlaer.  3roo.  21rod.  4ft.  139in.  the  2d,  99acr. 
4fyd.  ;Vthe  3d,  97acr.  24per.  8ft.  99in. ;  how  much  then, 
had  the  fourth  ? 

Ans.  244ac.  Iroo.  29per.  23£yds.  3ft.  48in. 


COMPOUND    SUBTRACTION.  59 

3.  A  man  had  1000  acres  of  land,  which  he  divided 
among  his  three  sons  ;  giving  Abraham  156ac.  3roo.  14 
rod.  27yds,  8ft.  125in. ;  Isaac  as  much  again ;  and  Jacob 
the  rest ;  pray  how  much  had  Jacob  ? 

Ans.  529ac.  Iroo.  85per.  6fyd.  57in. 

CUBIC  OR  SOLID  MEASURE. 

T.  rou.  ft.  inches.  T.  hewn.  ft.  Cord.  ft. 

64    37      1141  802    26  31     39 

17    19      1400  204     31  9  107 


APPLICATION. 

1.  Agreed  with  Noah  Nott  for  134  tons,  20ft.  of  round 
timber  T  he  has  drawn  99tons,  39ft.  1701m. ;  how  much 
more  must  he  haul  to  complete  the  contract  ? 

Ans.  34tons,  20ft.  27in. 

2. 1  bought  94tons,  25ft.  1600in.  of  hewn  timber,  of 
which  I  have  sold  F.  Francis  54tons,  45ft.  1709m. ;  and 
Saul  Swift  agrees  to  take  the  rest ;  how  much  less  will  be 
the  quantity  taken  by  Swift  than  that  taken  by  Francis  ? 

Ans.  15tons,  16ft.  90in. 

DRY  MEASURE. 

ChaL     bus.    pk.  Bus.  pic.  gal.  qt.  pt. 

103       17       1  62     1     0    2      1 

94       31       2  15     2     0     3      1 


APPLICATION. 

1.  A  merchant  contracted  for  6000  bushels  of  wheat, 
which  was  shipped  on  board  of  two  vessels  ;  one  arrived, 
and  brought  3000bush.  2pk.  Igal.  3qt.  Ipt. ;  the  other  nev- 
er came  into  port ;  how  much  less  was  lost  than  was  saved? 

Ans.  Ibus.  Ipk.  Igal.  3qts. 

2.  Having  199  chaldrons  of  coals,  I  sell  George  Green 
99chal.  34  bus.  Ipk.,  and  Mark  Mann  the  rest ;  how  much 
less  has  Mark  than  George  ?  Ans.  32bus.  2pk. 


80  DECIMAL    FRACTIONS. 

mal  point,  in  the  sum,  directly  under  the  decimal  points 
of  the  numbers  which  have  been  added. 


EXAMPLES. 


124,6201  3741,21 

5,92  374,646 

17,1174  8,46 

305,2165  52117,42 

2,71  91,5 


455,5840 


In  this  first  example,  the  sum  is  455  integers,  or  pounds, 
and  yVVb-  parts  of  a  unit  or  pound  ;  or  it  is  455  units,  and 
5  tenth  parts,  8  hundredth  parts,  and  4  thousandth  parts 
of  a  unit,  or  1  ;  the  cipher  at  the  right  of  the  decimal 
places  does  not  affect  the  value  of  the  other  figures,  and 
it  is,  therefore,  thrown  away. 

Hence  we  may  observe,  that  decimals,  and  Federal 
Money,  are  subject  to  one  and  the  same  law  of  notation, 
and,  consequently,  of  operation. 

For  since  1  dollar  is  the  money  unit,  and  a  dime  being 
the  tenth,  a  cent  the  hundredth,  and  a  mill  the  thousandth 
part  of  a  dollar,  it  is  evident  that  any  number  of  dollars, 
dimes,  cents,  and  mills,  is  simply  the  expression  of  dol- 
lars, and  decimal  parts  of  a  dollar:  thus,  11  dollars,  6 
dimes,  5  cents=ll,65  or  HT6^dol.  &c. 


3.  What  is  the  sum  of  276,+39,213+72014,9+417,+ 
5032,  and+2214,298  acres  ?         Ans.  79993,411  acres. 

4.  What  is  the  sum  of   ,014+,9816+,32  +  ,  15914  + 
,72913+,  0047  gallons  ?  Ans.  2,20857gal. 

5.  What  is  the  sum  of  27,148+918,73+14016,+294304, 
+7138,  and  +221,7  bushels  ?       Ans,  316625,578bus. 

6.  Add  the  following  sums  of  dollars  together,  viz. 
$12,34565+7,89l+2,34+14,+0011. 

Ans.  $36,57775  or  $36,  5di.  7cts. 


7.  To  9,999999  miles  add  one  millionth  part  of  a  mile-.. 

Ans*  10rail.es*. 


DECIMAL    FRACTIONS.  81 

SUBTRACTION  OF  DECIMALS. 

RULE. — Set  the  less  number  under  the  greater  in  the 
same  manner  as  in  addition  ;  then  subtract  as  in  whole 
numbers,  and  place  the  decimal  point  in  the  remainder 
directly  under  the  other  points. 

EXAMPLES. 

$.  £. 

612,32  16,279 

51,0942  8,0917 


561,2258 

4.  From  ,9173ft.  subtract  ,2138.        Ans.  ,7035  foot. 

5.  From  $2,73  subtract  $1,9185.  Ans.  $,8115. 

6.  Subtract  91,713acres  from  407.    Ans.  315,287ac. 

7.  What  is  the  difference  between  67tons  and  ,92  of  a 
ton  ?  Ans.  66,08  tons. 

8.  From  1  league  subtract  the  millionth  part  of  itself. 

Ans.  ,999999  league. 

MULTIPLICATION   OF  DECIMALS. 

RULE. — Place  the  factors,  (whether  mixed  numbers,  or 
pure  decimals,)  and  multiply  them,  as  in  whole  numbers  ; 
and  from  the  product,  towards  the  right  hand,  point  off 
as  many  figures  for  decimals  as  there  are  decimal  places 
in  the  factors.  But  if  there  be  not  so  many  figures  in 
the  product,  prefix  ciphers'  to  supply  the  defect. 

* 
EXAMPLES. 

1.  2. 

21,41  yards.  ,2616  ft. 

25,9  shillings.  ,154  doll. 


19269  10464 

10705  13080 

4282  2616 


554,519  shill.  Ans.  ,0402864  doll.  Ans. 


62  COMPOUND    MULTIPLICATION. 

AVOIRDUPOIS  WEIGHT. 

Ton.  cwt.  qr.  ft.  ft.  oz.  dr. 

1       17     3    23  21    11  15 

5  6 


Here  25ft.  a  qr. 


APPLICATION. 

1.  What  is  the  weight   of  6  barrels  of   sugar,   each 
weighing  Icwt.  3qrs.  20ft,.  1         Ans  llcwt.  2qrs.  8ft. 

2.  What  is  the  weight  of  12  hogsheads  of  sugar,  each 
13cwt.  2qrs.  23ft.,  25ft.  a  qr,  1  Ans.  164cwt.  3qrs.  1ft. 

3.  What  is  tlie  weight  of  6  chests  of  tea,  each  weigh- 
ing 3cwt.  2qrs.  9ft.,  25ft.  a  qr.  ?  Ans.  21cwt.  2qr.  41b. 


CLOTH  MEASURE. 

Yds.  qr.  na.         E.FL  qr.  na.         E.E.  qr.  na. 
313  91     a       1  61      4     3 

789 


APPLICATION. 

1.  What  number  of  yards  is  in  8  pieces  of  broadcloth, 
each  32yds.  3qrs.  Ina.  ?  Ans.  260yds.  2qrs. 

2.  If  8  E.  £.  3qrs.  3na.  of  broadcloth  will  make  a  suit 
of  clothes;  how  much  of  the  same  cloth  will  make   12 
similar  suits  7  Ans.  105  £.  Eng. 

LONG  MEASURE. 

Deg.  mi.  fur.  pol.  ft.  in.  bar.      Mi.  far.pol.yds.  ft. 
5      31     3     27    14   8      2         7     5      21    5     1 

10  11 


APPLICATION. 

1.  How  far  will  a  man  travel  in  7  days,  if  he  go3lmi. 
31  pol.  6ft.  6in.  every  day  ? 

Ans.  217mi.  6fur.  I9pol.  1 2ft.  6in. 

2.  If  in  a  race,  a  horse  move  14ft.  7in.  2bar.  at  every 
bound,  and  take  2  bounds  in  every  second,  what  course 
will  he  run  over  in  12  seconds  ?          Ans.  1 17yds.  4in. 


COMPOUND    MULTIPLICATION.  63 

LAND  OR  SQUARE  MEASURE. 
Ac.  roo.  per.  yd.  ft.  in.  Ac.  roo.  per. 

15     3      19     23    7    72  39     1     37 

8  9 


APPLICATION. 

1.  In  9  fields  each  containing  14  acres,  1  rood,  and  25 
perches,  how  many  acres  ?      Ans.  129ac.  2roo.  25per. 

2.  If  a  man  divide  his  farm  among  his  seven  sons,  and 
give  each  51  ac.  31  per.  8ft.  how  much  does  the  farm  con- 
tain ?  Ans.  358ac.  Iroo.  17per.  6yd.  2ft. 

CUBIC  OR  SOLID  MEASURE. 
T.rou.  ft.  in.  T. hewn,  ft.  Cord.  ft. 

1     39  845  24      49  12  124 

5  89 


APPLICATION. 

1 .  Bought  5  boat  loads  of  round  timber,  each  of  which 
contained  3tons,  28ft.  llllin. ;  what  is  the  whole  quan- 
tity 1  Ans.  18tons,  23ft.  371in. 

2.  In  six  parcels  of  wood,  each  containing  5cords,  96ft, 
how  many  cords  1  Ans.  34£cords. 

DRY  MEASURE. 

Chal.  bus.  pic.  Bus.  pk.  gal.  qt.  pt. 

54    35    2  93121 

6  7 


APPLICATION. 

\ .  There  were  six  wagons  loaded  with  coals,  each  of 
which  contained  Ichal.  8bus.  3pk. ;  what  was  the  total 
quantity  ?  Ans.  7chal.  16bus,  2pk. 

2.  Nine  loads  of  wheat  were  bought  by  a  miller,  each 
of  which  contained  21bus.  Ipk.  Igal.  Iqt.  Ipt. ;  what  was 
the  total  quantity  ?  Ans.  192bus.  3pk.  Iqt.  IpL 


64  COMPOUND    MULTIPLICATTON. 

WINE  MEASURE. 

Tun.  lihd.  gal.  qt.  Hhd.  gal.  qt.  pt.  gill 

10     3      17     1  1       49  2    1      I 

11  12 


APPLICATION. 

1.  How  much  wine  in  9  casks,  each  containing  41gal. 
3qts^  Ipt.  ?  Ans.  376gals.  3qts.  Ipt. 

"2.  How  much  cider  may  be  put  into  eight  casks,  each 
of  which  will  hold  104gals.  2qts.  Ipt.  3gills  ? 

-  Ans.  837f  gals. 

TIME. 

Yr.  m.  w.  d.  Ji.  m.  s.  Yr.  days.  h.  m. 

4      12  2    5  10  10  10  1      224     5  40 

12  8 


APPLICATION. 

If  a  solar  year  contain  just  365  days,  5  hours,  48  min- 
utes, and  48  seconds,  what  time  is  equal  to  7  solar  years  ? 
Ans.  2556days,  16h.  4  1m.  36s. 

CASE  II.  —  If  the  multiplier  exceed  12,  multiply  suc- 
cessively by  its  component  parts,  as  in  Case  II  in  sim- 
ple Multiplication. 

EXAMPLES. 

1.  What  will  &lyds.  of  calico  come  to,  at  2s.  7£d.  per 
yard  ? 


3x7=21 


18  4J  price  of  7  yards. 
3 


Ans.  ,£2  15  1J-  price  of  21  yards. 

2.  What  will  18  yards  come  to,  at  £1   7s.  %£d.  per 
yard  1  Ans,  <£24  9s.  9d. 

3.  What  will  36  pair  of  shoes  come  to,  at  13s.  4d.  per 
pair?  Ans. 


COMPOUND    MULTIPLICATION.  65 


4.  What  will  49  yards  of  broadcloth  cost,  at  17s. 
per  yard  ?  Ans.  ,£43  Os. 

5.  A  gentleman  is  possessed  of  l£  doz.  silver  spoons, 
each  weighing  2oz.  I5dwt.  llgr. ;  2  doz.  teaspoons,  each 
Wdwt.  I4gr. ;  and  two  silver  tankards,  each21oz.  \5dwt. ; 
pray  what  is  the  weight  of  the  whole  ? 

Ans.  8/6.  lOoz.  2dwt.  6gr. 

6.  What  is  the  weight  of  42  tubs  of  butter,  each  16/6. 
Uoz.  126/r.,  251b.  a  qr.  ?         Ans.  7cwt.  10/6.  lloz.  Sdr. 

7.  How  many  yards  in  81  pieces  of  cloth,  each  7yds. 
3#rs.  Ina.  1  Ans.  632y*/s.  3qrs.  \na. 

8.  How  many  bushels  in  63  casks,   each  containing 
46ws.  3pks.  Igal.  1  Ans.  3076ws.  IgaL 

CASE  III. — If  the  multiplier  be  not  a  composite  num- 
ber, find  the  nearest  to  it,  either  greater  or  less ;  multi- 
ply by  the  component  parts  as  before,  and  for  the  odd 
parts  add  or  subtract  as  the  case  requires. 

EXAMPLES. 

1.  What  will  65£  yards  of  cloth  come  to,   at  <£!   14s. 
i.  yer  yard  ? 

£1  14  6£ 
8 

8x8=64       13  16  4  price  of  eight  yards. 

8 


110  10  8   price  of  64  yards. 
4)         1    14  6J  price  of  1  yard. 
8  7£  price  of  £yard.+ 

Ans.  ,£112  13  10  price  65 %  yards. 
2.  What  will  76  yards  cost,  at  14s.  9±d.  per  yard  ? 

<£0  14  94 
7x11=77—1=76          II 


8    2  8£  price  of  11  yards. 

7 


56  18  11£  price  of  77  yardg. 
Subtract          14    9£  price  of  1  yard. 

Ans.  ,£56    4    2  price  of  76  yards. 
F2 


66  COMPOUND    DIVISION. 

3.  What  will  183  gallons  wine  cost,  at  7s.  5d.  ? 
10x10=100 

10xS=80 

3=3     Then  100+80+3=183. 

Ans.  ,£67  17s.  3d 

4.  What  will  600  yards  of  cloth  cost,  at  <£!  2s.  7±d.  1 

10x10x6=600  Ans.  .£678  15s.  Odf. 

5.  If  a  man  travel  46m?".  7fur.  30pol   3yds.  2ft.  every 
day,  for  57  J  days,  how  far  will  he  go  ? 

Ans.  2700mi.  Cyfur.  23poL  lyd.  2ft.  6in. 

6.  If  a  farm  consist  of  43  lots,   and  each  lot  contain 
3ac.  3r0o.  3per.  3yds.  3ft.  3m. ;  how  large  is  the  farm  ? 

Ans.  162«c.  I3per.  22yds.  3ft.  129m. 

7.  In  51  loads  of  bark,  each  1  cord,  26  feet,  how  many 
cords  ?  Ans.  61cor.  46/fc.  or  61e0r.  2£ft.  bark  meas. 

8.  If  a  cargo  of  wine  consist  of  122  casks,  and  each 
cask  contain  51gal.  3qts.  Ipt.  how  many  tuns  in  all  ? 

Ans.  25tun,  28gal.  3qts. 

9.  If  a  person  waste  Ihour,  32min.  41sec.  every  day 
for  38  years,  how  much  time  does  he  lose  in  the  whole 
^period,  allowing  365£  days  to  the  year  ? 

Ans.  2yrs.  162da.  19h.  58rn. 


COMPOUND  DIVISION. 

j+ 

COMPOUND  DIVISION  teaches  to  find  how  often  one  num- 
ber is  contained  in  another  of  different  denominations. 

RULE.— Place  the  numbers  as  in  Simple  Division.  Be- 
gin at  the  left  hand,  and  divide  each  denomination  by  the 
divisor,  setting  the  quotients  under  their  respective  divi- 
dends ;  but  if  there  be  a  remainder  in  dividing  any  of  the 
denominations  except  the  lowest,  find  how  many  of  the 
next  lower  denomination  it  is  equal  to  ;  and  add  it  to  the 
number,  if  any,  which  was  in  this  denomination  before  ; 
divide  this  sum  as  usual,  and  thus  proceed,  until  the 
whole  is  finished. 

The  method  of  proof  is  the  same  as  in  Simple  Division. 


COMPOUND  DIVISION.  67 

CASE  I.— EXAMPLES. 

MONEY, 

£      s.       d.  £     s.     d.     qr.  £      s.     fl. 

8)39    11      6£        6)22    942  5)71     13  8f 

4     18  1U+ 


APPL1CATIOJV. 

1.  Divide  £67  10s.  9d.  by  7.     Ans.  £9  12s. 

2.  Divide  £S  equally  among  6  persons. 

Ans.  £1  6s.  8d. 

TROY  WEIGHT 

ft.  oz.  dwt.  ft.    oz.  dwt. 

4)13     1     15  6)1       0      0 


APPLICATION. 

If  2ft.  9oz.  5dwt.  12gr.  of  silver  be  wrought  into   a 
dozen  spoons,  what  will  each  weigh  ? 

Ans.  2oz.  15dwt.  llgr. 

AVOIRDUPOIS  WEIGHT. 

Cict.     qr.    ft  Ton.  act.     qr, 

8)75        1     12  8)8      0      0 


Here  25ffo  a  qr. 


APPLICATION. 

1.  Divide  13cwt.  Iqr.  12ft.  6oz.  lOdr.  by  11. 

Ans.  Icwt.  24ft.  9T7Tdr. 

2.  Divide  17cwt.  Iqr.  of  sugar  equally  among  6  persons, 
at  25ft  a  qr.  ?  Ans.  2cwt.  3qrs. 

CLOTH  MEASURE. 

Yd.     qr.     na.  E.E.     qr.     na. 

5)31      23  9)1         00 


68  COMPOUND    DIVISION. 

K 

APPLICATION. 

1.  If  8  equal  pieces  of  cloth  contain  260yds.  2qrs.  what 
does  each  piece  contain  1  Ans.  32yds.  2qr.  Ina. 

2.  If  105  E.  Eng.  will  make  12  suits  of  clothes,  what 
does  it  take  for  one  suit  ?  Ans.  8E.  E.  3qr.  3na. 

LONG  MEASURE. 

Deg.  mi.  fur.  pol.  ft.  in.  bar.          Mi  fur.  pol.  yd.  ft. 
10)6      31     3    28     14    8     2         11)7    5      21     5     1 


APPLICATION. 

1.  If  a  man  travel  217m iles,  5fur,  I9pol.  12ft.  6in.  in 
7  days ;  how  far  does  he  go  a  day  1 

Ans.  31  mi.  31pol.  6ft.  6in. 

2.  If,  in  a  race,  a  horse  go  over  a  course  of  117yds.  4in. 
in  12  seconds,  how  far  does  he  move  in  one  second  ? 

Ans.  9yds.  2ft.  3in.  Iban 

LAND  OR  SQUARE  MEASURE. 

Ac.  roo.  per.  yd.  ft.  in.  Ac.  roo.  per. 

8)15      3    20      87    72  9)39      1      37 


APPLICATION. 

1.  If  nine  fields  of  equal  extent,  contain  129ac.  2roo. 
25per.  what  does  one  of  them  measure  ? 

Ans.  14ac.  Iroo.  25per. 

2.  If  a  man  divide  his  farm  of  35Sac.  Iroo.  17per.  6yd. 
2ft.  in  equal  portions  between  seven  sons,  what  does  each 
have  1  Ans.  51ac.  3lper.  Sft. 

CUBIC  OR  SOLID  MEASURE. 

T.rou.ft.     in.  T.  hew.  ft.  Cord.  ft. 

5)1       39    845          8)24        49          9)12     124 


COMPOUND    DIVISION.  69 

APPLICATION. 

1.  Bought  I8tons,  23ft.  371in.  of  round  timber,  which 
was  in  five  boats,  and  each  contained  a  like  quantity; 
how  much  did  one  boat  contain  1 

Ans.  3tons,  28ft.  llllin. 

2.  In  six  equal  parcels  of  wood  I  have  34^cords ;  what 
is  in  each  parcel.  Ans.  5cords,  96ft. 

DRY  MEASURE. 

Chal  bus.  pk.  Bus.  pic.  gal  qt.  pt. 

6)54       35     2  7)9     3       1      2     I 


APPLICATION. 

1.  If  in  nine  equal  loads  of  corn  there  be  192bus.  3pk. 
Iqt.  Ipt.,  what  is  there  in  one  load  1 

Ans.  21bus.  Ipk.  Igal.  Iqt.  Ipt. 

2.  Six  wagons  equally  loaded  drew  to  market  7chal. 
16bus.  2pk. of  coals  ;  how  much  did  one  bring  ? 

Ans.  Ichal.  8bus.  3pk. 

WINE  MEASURE. 

Tuns.hhd.  gal  qt.  Hhd.  gal.  qt.  pt.  gill. 

11)19    3      17     1  12)1      49    2    1      1 


APPLICATION. 

1.  In  nine  equal    casks,  I  have   376gal.   3qt.    Ipt.   of 
wine  ;  how  much  is  in  one  cask  ? 

Ans.  41gaL  3qts.  Ipt. 

2.  I  have  put  into  8  hogsheads  837Jgal.  of  cider,  each 
being  filled  alike  ;  how  much  is  in  each  cask  ? 

Ans.  104gal.  2qts.  Ipt.  3gills. 

TIME. 

Yr.  m.  w.  d.  7i.  m.  s.  Yr.  d.  h.  m. 

12)4     12  2    5  10  10  10  5)4  224  5  40 


70  COMPOUND    DIVISION. 

APPLICATION. 

If  in  seven  solar  years,  there  be  just  2556days,   16b. 
41m.  36s.,  what  is  the  length  of  one  solar  year  ? 

Ans.  365days,  5h.  48m.  48s. 

CASE  II. — If  the  divisor  exceed  12,  divide  continually 
by  its  component  parts,  as  in  Simple  Division,  CASE  III. 

EXAMPLES. 

1.  Divide  <£37  16s.  equally  among  24  men. 

'£.     s.     d. 
6)37    16    0 

4)6     6     0 
£1     11  6  Ans. 


2.  If  20  gallons  of  wine  cost  £7  5s.  lOd.  what  is  it 
per  gallon  ?  Ans.  7s.  3£d. 

3.  Bought   3doz.  of  silver  spoons,   which,  together, 
weighed  4]fo.  8oz.  12gr. :  how  much  silver  did  each  spoon 
contain  ?  Ans.  3oz.  4dwt.  llgr. 

4.  Divide  a  hhd.  of  sugar,   weight    12cwt.   3qrs.   7|fc 
equally  among  16  men,  25ft.  a  qr.  Ans.  3qrs.  5Jfo.  2oz. 

5.  Divide  43yds.  Iqr.  Ina.  of  crape  among  33  persons. 

Ans.  lyd.  Iqr.  Ina. 

6.  If  a  person  travel  171eag.   Imi.  4fur.  21pol.  in  21 
hours,  what  was  the  average  distance  an  hour  t 

Ans.  2mi.  4fur.  Ipol. 

7.  Divide  1000  acres  of  land  equally   among  99  per- 
sons. Ans.  lOac.  16£fper. 

8.  Divide  500  cords  of  bark  in  equal  parts  among  108 
persons.     Ans.  4cor.  80T6Q4¥ft.  or  4cor.  S^yft.  bark  meas. 

9.  Divide  168bus.   Ipk.    IgaL   2qts.   of   corn  equally 
among  35  persons.  Ans.  4bus.  3pk.  2qt. 

10.  Divide  4^gal.  of  wine  equally  among  144  soldiers. 

Ans.  Igill  each. 

CASE  III. — If  the  divisor  be  not  a  composite  number, 
divide  as  in  long  division. 


COMPOUND    DIVISION.  71 

EXAMPLES. 
'   1.  Divide  ,£391  17s.  6£d.  equally  among  46  men. 

£       s.      d.  £     s.      d. 
46)391    17     6£(8  10    4£  Ans. 
368 

23 
20 

46)477(105. 
46 

17 
12 

46)210(4d. 
184 

26 
4 

46)105(2^5. 
92 

13 

2.  If  263  bushels  of  wheat  cost  ^86  11s.  5d.  what  is 
it  per  bushel  1 

£      s.       d. 

263)86     11      5(<£0  6s.  7d.  Ans. 
20 

263)1731  (6s.  <fcc. 

3.  Divide  27tons,  13cwt.  2qrs.  of  iron  equally  among 
34  men,  251b.  a  qr. 

T.  cwt.  qr.T.  cwt.  qr.  ft. 
34)27     13    2(0     16    1      2  and  32  rem.  or  ff. 
20 

34)553(16cwt.  <fcc.  Ans.  16cwt.  Iqr.  21b. 

4.  If  461b.  of  indigo  cost  ^53  10s.  6d.,  what  i«  it  per 
pound  ?  Ans.  £\  35. 


72  COMPOUND    DIVISION. 

5.  If  263  bushels  of  wheat  cost  $287,  97cts.  3mills, 
what  is  it  per  bushel  ?  Ans.  $1,  9cts.  4mills..{- 

6.  If  37  thousand  of  boards  come  to  $203,50cts.,  what 
is  one  thousand  worth  ?  Ans.  $ 5,50cts. 

7.  Divide  120  months,  2w.  3d.  5h.  20m.  by  111. 

Ans.  Imo.  2d.  lOh.  l2T6T8Tm. 

8.  A  privateer  takes  a  prize  worth   $12465,  of  which 
the  owner  takes  one-half,  the  officers  one-fourth,  and  the 
remainder  is  equally  divided  among  the  sailors,  who  are 
125  in  number  ;  how  much  is  each  sailor's  part  ? 

Ans.  $24,93cts. 

9.  Three  merchants,  A,  B,  and  C,  have  a  ship  in  com- 
pany ;  A  has  f,  B  f ,  and  C  £;  they  have  received  for 
freight  ,£228  16s.  8d. ;  it  is  required  to  divide  it  among 
the  owners,  according  to  their  respective  shares  ;  pray, 
can  you  do  it  ?  .        (  A's  share  <£143  5d.  B's  ,£57 

s'\      4s.  2d.      C's  ^28  12s.  Id. 

10.  A  privateer  having  taken  a  prize  worth  $6850,  it 
is  divided  into  one  hundred  shares  ;  the  captain  takes  1 1  ; 
2  lieutenants,  each  5  ;   12  midshipmen,  each  2  ;  and  the 
remainder  is  to  be  equally  divided  among  the  sailors,  who 
are  105  in  number; — pray,  can  you  settle  the  matter  ? 

.        i  Captain's  share  $753,50cts. ;  a  lieut's  $342,50cts. ; 
s*  \  a  midshipman's  $137;  and  a  sailor's  $35,88TV\cts. 

11.  Divide  the  sum  of  50  eagles,  50  dollars,  50  dimes, 
50  cents,  and  60  mills  among  17  men  ;  and  give  the  first 
12  cents  more  than  the  second,  the  second  12  cents  more 
than  the  third,  arid  so  on  to  the  last; — what  will  the  sev- 
enteenth man's  share  be  1  Ans.  $31,72cts. 

NOTE. — It  is  customary  among  appraisers  of  property, 
arbitrators  or  referees,  and,  in  some  cases,  the  method  is 
resorted  to  even  by  juries,  where  they  cannot  agree  in 
their  estimates  and  verdicts,  to  take  the  amount  of  the 
sums  which  they  severally  agree  to  award,  and  divide  that 
amount  by  the  number  of  which  they  consist ;  the  quo- 
tient is  their  average  judgment,  which  goes  for  their  esti- 
mate, decision,  or  verdict. 

EXAMPLES. 

1.  In  appraising  a  certain  property,  A  called  the  value 
$100,  B,  $140,  C,  $80,  and  D,  $150  ;  but  as  they  could 


DUODECIMALS. 


73 


not  agree  in  their  estimates,  they  determined  to  take  their 
average  judgment,  and  let  that  be  the  value  ;  what  was 
the  value  in  that  case  ? 

A        $  100 

B  140 

C  80 

D  150 

4  )470 

$117,50c.  Ans. 

2.  Four  persons  appraising  a  building,  A  valued   it  at 
$550,50c.,  B,  at  $480,50c.  €,   at   $590,50c.   and  D,   at 
$600  ;  what  was  the  mean  value?         Ans.  $555,37^c. 

3.  Seth   Strong,    Luke  Locke,   Mark  Mills,   Charles 
Church,  and  John  Jones,  were  appointed  to  appraise  the 
ship  Ocean  ;  Strong's  estimate  was  $5500,  Locke's  $7000, 
Mill's    $6666,    Church's    $8444,    and    Jones's,     $5000; 
what  was  the  mean  judgment  ?  Ans.  $6522. 


DUODECIMALS. 

DUODECIMALS  chiefly  regard  feet  and  inches.  They 
are  so  called,  because  they  decrease  by  twelve  from  the 
place  of  feet  towards  the  right  hand. 

Inches  are  sometimes  called  primes,  and  marked  thus 
(')  ;  the  next  division  is  called  parts  or  seconds,  and 
marked  (*)  ;  the  next  thirds,  and  marked  (m)  ;  &c. 

MULTIPLICATION  OF  DUODECIMALS. 

RULE.  —  Under  the  multiplicand  write  the  corresponding 
denomination  of  the  multiplier.  Multiply  each  term  in 
the  multiplicand,  beginning  at  the  lowest,  by  the  highest 
denomination  in  the  multiplier  ;  and  write  each  result 
under  its  respective  term  ;  observing  to  carry  a  unit  for 
every  12  from  each  lower  place  to  its  next  higher. 

In  the  same  manner  multiply  all  the  multiplicand  by 
the  next  highest  denomination  in  the  multiplier;  and  set 
G 


74  DUODECIMALS. 

the  result  of  each  term  removed  one  place  to  the  right 
hand  of  those  in  the  multiplicand. 

Proceed  in  like  manner  with  the  remaining  denomina- 
tions, and  the  sum  of  all  the  lines  will  be  the  product  re- 
quired. 


EXAMPLES. 

1.  Multiply  4  feet  2  inches  by  3  feet  5  inches. 

Ft.   ' 

The  2  in  the  product     4    2 
is  not  2  inches,  but  235 

twelfths  of  a  square  foot, 

or  24  inches,  &c.  12     6 

1     8     10" 


14    2     10  Ans.  2ft.  2' 10". 


2.  Multiply  10  feet  11  inches  by  7  inches. 

10     11 

7 

645  Ans.  6ft.  4'  5". 

3.  What  is  the  content  of  a  bale  6  feet  5'  long ;  4  feet 
3'  high,  and  3  feet  10'  wide  ? 

Ft.  ' 
6  5 
4  3 

25      8    H 
173 


27     3  3 
3  10 

81     9  9    m 

22    8  8    6 

104    6  5  "6  Ans.  104ft.  6'  5"  6W. 


VULGAR    FRACTIONS.  75 

4.  What  is  the  content  of  a  marble  slab  4ft.  ?'  S"  wide 
and  5ft.  6'  long?  Ans.  25ft.  6'  2". 

5.  Multiply  7ft.  8'  6"  by  10ft.  4'  5". 

Ans.  79ft.  11'  0"  6'"  6"". 

6.  Multiply  44ft.  2'  9"  2'"  4""  by  2ft.  10'  3". 

Ans.  126ft.  2'  10"  8'"  10""  11'"". 

7.  How  many  square  feet  in  a  board  25  feet  6  inches 
long,  and  1  foot  §  inches  wide  1 

Ans.  31  feet  10£  inches. 

8.  How  many  cubic  feet  in  a  stick  of  timber   12  feet 
10'  long,  1  foot  7'   wide,  and  1  foot  9'  thick  1 

Ans.  35ft.  6'  8V  6'". 

9.  How  many  cubic  feet  of  wood  in  a  load  7  feet   10' 
long,  3  feet  11'  wide  and  3  feet  6'  high  1 

Ans.  107ft.  4'  7". 

10.  The  length  of  a  room  being  20ft.,  its  width  14ft.  6', 
and  height  1  Oft.  4'  ;  how  many  yards  of  painting  are  in  it, 
deducting  a  fire-place  of  4ft.  by  4ft.  4',  and  two  windows, 
each  6ft.  by  3ft.  2'  ?  Ans.  73^-yards. 

11.  Required  the  solid  contents  of  a  wall  53ft.  6'  long, 
10ft.  3'  high,  and  2ft.  thick.  Ans.  1096ft.  9'. 


FRACTIONS. 

FRACTIONS,  or  broken  numbers,  are  expressions  for  any  t 
assignable  parts  of  a  unit,  or  whole  number,  and,  gene- 
rally, are  of  two  kinds,  viz. 

VULGAR  AND  DECIMAL. 

A  Vulgar  Fraction  is  represented  by  two  numbers, 
placed  one  above  the  other,  with  a  little  line  drawn  be- 
tween them  ;  thus,  £,  |,  &c.  signify  three-fourths,  five- 
eighths,  &c. 

The  figure  above  the  line,  is  called  the  numerator,  and 
the  one  below  the  line  the  denominator. 

(  5  Numerator. 
Thus  J  — 

(  8  Denominator. 
The  denominator  (which  is  the  divisor,  in  division,) 


76  VULGAR    FRACTIONS. 

shows  how  many  parts  the  unit  or  integer  is  divided  into  ; 
and  the  numerator  (which  is  the  remainder  after  division,) 
shows  how  many  of  those  parts  are  meant  by  the  fraction  : 
A  fraction  is  said  to  be  in  its  least  or  lowest  terms, 
when  it  is  expressed  by  the  least  numbers  possible  ;  thus 
|  when  reduced  to  its  lowest  terms,  will  be  ^,  and  T9^  when 
expressed  by  the  least  numbers  possible,  will  be  f  ;  and 
so  of  any  others. 

PROBLEM  I.    , 

To  abbreviate  or  reduce  fractions  to  their  lowest  terms. 

RULE.  —  Divide  the  terms  of  the  given  fraction  by  any 
number  which  will  divide  them  without  leaving  a  remain- 
der, and  the  quotients  divide  again  in  like  manner  ;  and 
so  on  till  it  appears  that  there  is  no  number  greater  than 
1,  which  will  any  longer  divide  them  \  and  then  the  frac- 
tion will  be  in  its  lowest  terms. 

EXAMPLES. 

1.  Reduce  f-|-§  to  its  lowest  terms. 


=f  the  answer. 

2.  Reduce  -J-f  f  to  its  lowest  terms.  Ans.  ^. 

3.  Reduce  f  ^f  to  its  lowest  terms.  Ans.  \. 

4.  Reduce  /Y\-  to  its  lowest  terms.  Ans.  ^. 

5.  Abbreviate  ff  as  much  as  possible.  Ans.  \^. 

0.  Bring  ff^  to"the  least  terms  possible.  Ans.  ff  . 

7.  Reduce  ^ff  to  its  lowest  terms.  Ans.  f. 

8.  Abridge  •££%  as  low  as  you  can.  Ans.  £. 

9.  Put  \\±  into  its  least  terms.  Ans.  £%. 
10.  Let  I  iff  be  expressed  in  its  least  terms.  Ans.  f. 

PROBLEM  2. 

To  find  the  value  of  a  fraction  in  the  known  parts  of 

the  integer,  as  of  coin,  weight,  measure,  fyc. 
RULE.  —  Multiply  the  numerator  by  the  common  parts 
of  the  integer,  and  divide  by  the  denominator,   and  the 
several  quotients  set  in  one  line,  will  show  the  answer. 

EXAMPLES. 

1.  What  is  the  value  of  f  of  a  pound  sterling  1 


VULGAR    FRACTIONS.  77 

Numer.  2 

20  shillings  in  a  pound. 

Denom.  3)40(l3s.  4d.  the  Ans. 
3 

10 
9 

1 
12  pence  in  a  shilling. 

3)12(4d. 
12 

2.  What  is  the  value  of  ±%  of  a  pound  sterling? 

Ans.  18s.  5d.  2T2oqrs. 

3.  Reduce  f  of  a  shilling  to  its  proper  quantity. 

Ans.  4d.  3^qrs. 

4.  What  is  the  value  of  f  of  12s.  6d.  7     Ans.  4s. 

5.  What  is  the  value  of  ff  of  a  pound  Troy  7 

Aris.  9oz. 

6.  How  much  are  T9T  of  a  hundred  weight  ? 

Ans.  3qrs.  7fc.  10T2Toz. 

7.  What  is  the  value  of  f  of  a  mile  ? 

Ans.  6fur.  26pol.  lift. 

8.  How  much  are  £  of  a  cwt.,  at  25fb  to  a  qr.  ? 

Ans.  3qrs.  Sffe.  12oz.  7|dr. 

9.  Show  the  proper  quantity  of  f  of  an   ell  English  I 

Ans.  2qrs.  3jna. 

10.  How  much  are  %  of  a  hhd.  of  wine  ? 

Ans.  54  gallons. 

11.  What  is  the  value  of  T%  of  a  day  I 

Ans.  iGh.  36m.  55T5TrS. 

12.  What  is  the  value  of      of  a  dollar  1  Ans.  80cts. 


PROBLEM   3* 

To  reduce  any  given  quantity   to  the  fraction  of  any 
greater  denomination  of  the  same  kind. 

RULE.  —  Reduce  the  given  quantity  to  the  lowest  term 
mentioned,  (as  in  reduction  of  Moneyr  Weights,, 
G2 


78  VULGAR    FRACTIONS. 

nres,  &c.)  for  a  numerator  ;  then  reduce  the  integral  part 
to  the  same  term,  for  a  denominator  1  place  the  former 
over  the  latter,  and  they  will  be  the  fraction  required, 
which,  if  it  be  not  in  its  lowest  terms,  must  be  reduced 
by  Problem  1. 

EXAMPLES. 

1.  Reduce  13s.  6d.  2qrs.  to  the  fraction  of  a  pound* 

5.  d.  qrs. 

20  shill.  integral  part.  136    2  given  sum. 
12  12 

240  162 

4  4 

960  Denominator,.        650  Numerator. 

Ans.      £= 


2.  What  part  of  a  hundred  weight  are  3qrs. 

Ans,  T^= 


3.  What  part  of  a  yard  are  3qrs,  3n.a.  T  Ans.  | 

4.  What  part  of  a  pound  sterling  are  13s.  4d. 

Ans. 

5.  What  part  of  a  civil  year  are  3  weeks,  4  days  7 

Ans.  32g5?=T 

6.  What  part  of  a  mile  are  6fur.  26pol.  3yds.  2f.. 

fur.  pal.  yds.  ft.    feet. 
6     26     3     2==44QO  Num. 

a  mile  =5280  Denom.          Ans. 


7.  Reduce  7oz.  4dwt.  to  the  fraction  of  a  }fr.  Troy. 

Ans.  f  . 

8.  What  part  of  am  acre  are  2  roods,  20  perches  1 

Ans.  •§. 

9.  Reduce  54  gallons  to  the  fraction  of  a  hhd.  of  wine. 

Ans.  -|. 

10.  What  part  of  a  hogshead  of  wine  are  9  gallons  ? 

Ans.  ^. 

11.  What  part  of  a  pound  Troy  are  lOoz.  lOdwt.  lOgr.  1 

Ans.  &%. 

12.  What  part  of  a  dollar  are  80cts.  T  Ans.  |-. 


DECIMAL    FRACTIONS.  7*J 

DECIMAL  FRACTIONS. 

A  DECIMAL  is  a  fraction  whose  denominator  is  a  unit, 
with  as  many  ciphers  annexed  to  it  as  the  numerator  has 
places  ;  and  is  usually  expressed  by  writing  the  numer- 
ator only  with  a  point  before  it  called  the  separatrix. 
Thus  T5<y,  T2<£p  T%3o6<T>  are  decimal  fractions,  and  are  ex-- 
pressed by  ,5  ,25  ,236  respectively. 

The  place  of  a  figure  in  decimals,  as  in  whole  numbers, 
determines  its  relative  value  :  That  in  the  first  place  next 
the  separatrix  is  10th  parts;  that  in  the  second,  100th 
parts,  &c.  decreasing  in  the  same  tenfold  proportion  to 
the  right  hand,  as  whole  numbers  increase  decimally  from 
units  to  the  left  hand. 

Ciphers  placed  at  the  right  hand  of  decimals,  make  na 
alteration  in  their  value  ;  for  ,5  ,50  ,500,  &c.  are  decimals 
of  the  same  vajue,  being  each  equal  to  ^ ;  but.  if  placed 
at  the  left  hand,  the  value  of  the  fraction  is  decreased  in 
a  tenfold  proportion  for  every  cipher  prefixed  ;  thus  ,5 
,05  ,005,  &<c.  are  5  tenth  parts,  5  hundredth  parts,  and  5 
thousandth  parts  respectively.  In  the  following  Table, 
the  doctrine  is  exemplified  at  large. 

NUMERATION  TABLE,  OR  SCALE   OF  NOTA^ 
TION, 


S 

7 


03 


o 
G 


3 
O 

EH 


M 
5 


a 

03 


H 
4 


a 

B 
3 


a      .ti 

<U  3 


03 
P«. 


s 

CD 


2      1, 


as 


3 

M 
3 


03 
OH 


3 
C 

H 
4 


£3 
03 
CO 

3 
O' 

H 
XI 
5 


03 

P- 


3 
O 

H 


•-* 
03 


O 


6      7 


WHOLE    NUMBERS. 


DECIMALS. 


ADDITION  OF  DECIMALS. 

RULE. — Set  the  numbers  so  that  the  decimal  points 
may  stand  directly  under  each  other,  then  add  as  in  whole 
s,  carrying  one  for  every  ten,  and  place  the  deci- 


80  DECIMAL    FRACTIONS. 

mal  point,  in  the  sum,  directly  under  the  decimal  points 
of  the  numbers  which  have  been  added. 


EXAMPLES. 

•£  $. 

124,6201  3741,21 

5,92  374,646 

17,1174  8,46 

305,2165  52117,42 

2,71  91,5 


455,5840 


In  this  first  example,  the  sum  is  455  integers,  or  pounds, 
and  ^0%  parts  of  a  unit  or  pound  ;  or  it  is  455  units,  and 
5  tenth  parts,  8  hundredth  parts,  and  4  thousandth  parts 
of  a  unit,  or  1  ;  the  cipher  at  the  right  of  the  decimal 
places  does  not  affect  the  value  of  the  other  figures,  and 
it  is,  therefore,  thrown  away. 

Hence  we  may  observe,  that  decimals,  and  Federal 
Money,  are  subject  to  one  and  the  same  law  of  notation, 
and,  consequently,  of  operation. 

For  since  1  dollar  is  the  money  unit,  and  a  dime  being 
the  tenth,  a  cent  the  hundredth,  and  a  mill  the  thousandth 
part  of  a  dollar,  it  is  evident  that  any  number  of  dollars, 
dimes,  cents,  and  mills,  is  simply  the  expression  of  dol- 
lars, and  decimal  parts  of  a  dollar:  thus,  11  dollars,  6 
dimes,  5  cents=ll,65  or  HT^dol.  &c. 


3.  What  is  the  sum  of  276,+39,213+72014,9+417,+ 
5032,  and-f  2214,298  acres  ?         Ans.  79993,411  acres. 

4.  What  is  the  sum  of   ,014  +  ,9816+,32  +  ,  15914  + 
,72913+,  0047  gallons  ?  Ans.  2,20857gal. 

5.  What  is  the  sum  of  27,148+918,73+14016,+294304, 
+7138,  and  +221,7  bushels  ?       Ans,  316625,578bus. 

6.  Add  the  following  sums  of  dollars  together,  viz. 
$12,34565+7,891  +2,34+  14,-fOOll. 

Ans.  $36,57775  or  $36,  5di.  7cts.  7^mills. 

7.  To  9,999999  miles  add  one  millionth  part  of  a  mile* 


DECIMAL    FRACTIONS.  81 

SUBTRACTION  OF  DECIMALS. 

RULE. — Set  the  less  number  under  the  greater  in  the 
same  manner  as  in  addition ;  then  subtract  as  in  whole 
numbers,  and  place  the  decimal  point  in  the  remainder 
directly  under  the  other  points. 

EXAMPLES. 

$.          £.  yds. 

612,32        16,279          37, 
51,0942       8,0917          2,41 


561,2258 

4.  From  ,9173ft.  subtract  ,2138.        Ans.  ,7035  foot. 

5.  From  $2,73  subtract  $1,9185.  Ans.  $,8115. 

6.  Subtract  91,7 lucres  from  407.    Ans.  315,287ac. 

7.  What  is  the  difference  between  67tons  and  ,92  of  a 
ton  ?  Ans.  66,08  tons. 

8.  From  1  league  subtract  the  millionth  part  of  itself. 

Ans.  ,999999  league. 


MULTIPLICATION   OF  DECIMALS. 

RULE. — Place  the  factors,  (whether  mixed  numbers,  or 
pure  decimals,)  and  multiply  them,  as  in  whole  numbers  ; 
and  from  the  product,  towards  the  right  hand,  point  off 
as  many  figures  for  decimals  as  there  are  decimal  places 
in  the  factors.  But  if  there  be  not  so  many  figures  in 
the  product,  prefix  ciphers'  to  supply  the  defect. 

v 

EXAMPLES. 

1.  2. 

21,41  yards.  ,2616ft- 

25,9  shillings.  ,154  doll. 


19269  10464 

10705  13080 

4282  2616 


554,519  shill.  Ans.  ,0402864  doll.  Ans. 


82  DECIMAL    FRACTIONS. 

3.  Multiply  31,72rods  by  65,3.       Prod.  2071,316rods. 

4.  Multiply  ,62foot  by  ,04.  Prod.  ,0248foot. 

5.  Multiply  51,6yards  by  21.  Prod.  1083,6yards. 

6.  Multiply  ,051,£  by  ,0091.  Prod.  ,0004641«£. 
7  What  cost  6,21  yards  of  cloth,  at  2dols.  32cts.  5 

mills  per  yard  ?  Ans.  $14,  4d.  3c.  8^/wm. 

8.  Multiply  7,02  dollars  by  5,27  dollars. 

Ans.  36,9954dols.  or  $36,99cts.  5T\m. 

9.  Multiply  41dols.  25cts.  by  120dols.         Ans.  $4950. 

10.  Multiply  3dols.  45cts.  by  16cts. 

Ans.  ,5520  or  55cts.  2m. 

11.  Multiply  65cts.  by  ,09  or  9cts. 

Ans.  ,0585=5cts.  8£m. 

12.  Multiply  lOdols.  by  lOcts.  Ans.  $1. 

13.  Multiply  341,45dols.  by  ,007  or  7  mills. 

Ans.  $2,39+. 

NOTE. — To  multiply  Decimal  Fractions  by  10,  100, 
1000,  &c.  is  only  to  remove  the  separatrix  so  many  pla- 
ces towards  the  right  hand,  as  there  are  ciphers  in  the 
multiplier. 

EXAMPLES. 

Multiply  $64,674  by  $10.  Ans.  $646,74. 

Multiply  $3,2158  by  1000  cords.  Ans.  $3215,8. 


DIVISION  OF  DECIMALS. 

RULE. — Divide  as  in  whole  numbers  ;  and  observe  the 
following  rules  for  pointing  off  in  the  quotient. 

1.  Point  off  for  decimals  In  the  quotient  so  many  fig- 
ures, as  the  decimal  places  in  the  dividend  exceed  those 
in  the  divisor. 

2.  If  the  figures  in  the  quotient  are  not  so  many  as  the 
rule  requires,  supply  the  defect  by  prefixing  ciphers. 

3.  If  the  decimal  places  in  the  divisor  be  more  than 
those  in  the  dividend,  add  ciphers  as  decimals  to  the  divi- 
dend, until  the  number  of  decimals  in  the   dividend  be, 
equal  to  those  in  the  divisor,  and  the  quotient  will  be  in- 
tegers until  all  these  decimals  are  used.     And  in  case  of 
a  remainder,  after  all  the  figures  of  the  dividend  are  used, 
and  more  figures  are  wanted  in  the  quotient,  annex  ci- 


DECIMAL    FRACTIONS.  83 

pliers  to  the  remainder,  to  continue  the  division  as  far  as 
necessary. 

4.  The  first  figure  of  the  quotient  will  possess  the  same 
place  of  integers  or  decimals,  as  that  figure  of  the  divi- 
dend which  stands  over  the  unit's  place  of  the  first  pro- 
duct. 

EXAMPLES. 

1.  Divide  3424,6056  by  43,6. 

Divisor.  Dividend.  Quotient. 
43,6)3424,6056(78,546 

3052  Or,  What  is  potash  per 

ton,  when  43  tons,  12cwt. 

3726  cost  $3424,6056  ? 

3488  Ans.  $78,54c.  6m. 


2380 

2180 

2005 
1744 

2616 
2616 

2.  Divide  761,2  miles  by  2,1942  weeks. 

2,1942)761,2000(346,91  miles,  and  10078  rem. 

3.  Divide  7,735  rods  by  3,25  Ans.  2,38  rods. 

4.  Divide  3877875«£  by  ,675.  Ans.  5745000«£. 

5.  Divide  1835,78  tons  bjt  7,48.    Ans.  245,42tons.+ 

6.  Divide  ,55736  cord  by  48.       Ans.  ,01161cord.+ 

7.  Divide  7,13  acres  by  18.  Ans.  ,396acre.+ 

8.  Divide  246,1476dols.  by  603,25dols. 

Ans.  ,40736dol.+ 

9.  Divide  186513,239dols.  by  304,81dols. 

Ans.  6il,9dols.+ 

10.  Divide  l,28dols.  by  8,31  dols. 

Ans.  ,154=l5cts.  4m.+ 

11.  Divide  56cts.  by  Idol.  12cts.  Ans.Sdimes  or  50cts. 

12.  Divide  1  dollar  by  12  cents. 

Ans.  8,333dols.  or,  $8,  33cts.  3m. 


84  DECIMAL    FRACTIONS. 

13.  If  2lf  or  21,75  yards  of  cloth  cost  34,517dols., 
what  is  the  price  of  one  yard  1  Ans.  l,586dol.+ 

NOTE. — When  decimals  or  whole  numbers,  are  to  be 
divided  by  10,  100,  1000,  &c.  (viz.  unity  with  ciphers,) 
it  is  performed  by  removing  the  separatrix  in  the  divi- 
dend, so  many  places  towards  the  left  hand,  as  there  are 
ciphers  in  the  divisor. 

EXAMPLES. 


(  10  the  quotient  is  74,8dols. 
'< 


$748  divided  by  I  100        -        -        7,48dols. 
(  1000  -     ,748dol. 


REDUCTION  OF  DECIMALS. 

CASE  1. 

To  reduce  a  Vulgar  Fraction  to  its  equivalent  decimal. 

RULE.— Divide  the  numerator  by  the  denominator,  an- 
nexing as  many  ciphers  as  are  necessary  ;  and  the  quo- 
tient will  be  the  decimal  required. 

EXAMPLES. 

1.  Reduce  A-  to  a  decimal. 

24)5,0000(,20833+Ans. 

48 

200 
192 


80 

72 

80 

72 

8  Remainder. 

2.  Required  the  equivalent  decimal  expressions    for  |« 
and  f .  Ans.  ,25  ,5  and  ,75. 

3.  Reduce  f  to  a  decimal.  Ans.  ,375. 

4.  Reduce  ^  and  ff  to  decimals.  Ans.  ,04  and  ,407.+ 


DECIMAL    FRACTIONS.  85 

5.  Reduce  §f  arid  T|T  to  decimals. 

Ans.  ,88  and  ,00689.+ 

CASE  II. 

To  reduce  numbers  of  different  denominations  to  their  equiv- 
alent decimal  values. 

RULE  1. — Write  the  given  numbers  perpendicularly 
under  each  other  for  dividends,  proceeding  orderly  from 
the  least  to  the  greatest. 

2.  Opposite  to  each  dividend,  on  the  left   hand,   place 
such  a  number  for  a  divisor,  as  will  bring   it  to  the  next 
superiour  denomination,  and  draw  a  line  between  them. 

3.  Begin  with  the   uppermost,  and  write  the  quotient 
of  each  division,  as  decimal  parts,  on  the  right  hand   of 
the  dividend  next  below  it ;  and  so  on  until  they   are  all 
used,  and  the  last  quotient  will  be  the  decimal  sought. 

EXAMPLES. 

1.  Reduce  15s.  9fd.  to  the  decimal  of  a  pound. 
3, 


20 


9,75 
15,8125 


,790625  the  decimal  required. 

2.  Reduce  19s.  to  the  decimal  of  a  pound. 

Ans.  ,95. 

3.  Reduce  10s.  9d.  Iqr.  to  the  decimal  of  a  pound. 

Ans.  ,5385416.+ 

4.  Reduce  Id.  2qrs.  to  the  decimal  of  a  shilling. 

Ans.  ,125. 

5.  Reduce  lOoz.  18dwt.  16grs.  to  the  decimal  of  a  Jk 
Troy.  Ans.  ,911111.+ 

6.  Reduce  lOoz.  14drs.  to  the  decimal  of  a  hundred 
weight.  Ans.  ,0060686.+ 

7.  Reduce  3  rods  2^-feet,  6  inches  to  the  decimal  of  a 
mile.  Ans.  ,00994318.+ 

8.  Reduce  1  pint  to  the  decimal  of  a  gallon. 

Ans.  ,125. 

9.  Reduce  2  months,  2  weeks,  2  days  to  the  decimal 
of  a  year.  Ans.  ,197302.+ 

H 


86  DECIMAL    FRACTIONS, 

10.  Reduce  3s.  4d.  New-England  currency  to  the  de- 
cimal of  a  dollar.  Ans.  ,555555.+ 

CASE  III. 
To  reduce  any   number  of  shillings,  pence  and  farthings 

by  inspection  to  the  decimal  of  a  pound. 
RULE. — Write  half  the  greatest  even  number  of  shil- 
lings for  the  first  decimal  figure,  and  let  the  farthings  in 
the  given  pence  and  farthings  possess  the  second  and 
third  places  ;  observing  to  increase  the  second  place  by 
5,  if  the  shillings  be  odd  ;  and  the  third  place  by  one, 
when  the  farthings  exceed  12,  and  by  2,  when  they  ex- 
ceed 36. 

EXAMPLES. 

1.  Find  the  decimal  of  15s.  S£d.  by  inspection. 

,7=i  of  14s. 
,05  for  the  odd  shilling. 
,034=farthings  in  8£d. 
,001  for  excess  above  12. 

,785=decimal  required. 

2.  Find  by  inspection  the  decimal  of  12s.  6£d. 

Ans.  ,628. 

3.  Find  by  inspection  the  decimal  of  18s.  lO^d. 

Ans.  ,943. 

4.  Find  by  inspection,  and  add  together  the  decimal  of 
13s.  6d.,  9s.,  Is.  9d.,  5d.  f,  and  l^d.       Ans.  .£1,242.+ 

CASE  IV. 
Tojind  the  value   of  any  given   decimal  in  terms  of   the 

integer. 

RULE  1. — Multiply  the  decimal  by  the  number  of  parts 
in  the  next  less  denomination,  and  cut  off  as  many  places 
for  a  remainder  on  the  right  hand,  as  there  are  places  in 
the  given  decimal. 

2.  Multiply  the  remainder  by  the  parts  in  the  next  in- 
feriour  denomination,  and  cut  off  for  a  remainder  as  be- 
fore. 

3.  Proceed  in  this  manner  through  all  the  parts  of  the 
integer,  and  the  several  denominations,  standing  on  the 
left  hand,  will  make  the  answer. 


DECIMAL    FRACTIONS.  87 

EXAMPLES. 

1.  What  is  the  value  of  ,74*26  of  a  pound? 
,7426 
20 

s.  14,8520 

12 

d.  10,2240 
4 


,8960     Ans.  14s.  10|d.+ 

2.  What  is  the  value  of  ,384  of  a  shilling?  Ans.  4£.+ 

3.  What  is  the  value  of  ,6725cwt.,  at  251b.  a  qr. 

Ans.  2qrs.  171b.  4oz. 

4.  What  is  the  value  of  ,61  of  a  tun  of  wine  ? 

Ans.  2hhds.  27gals.  2qts.  lpt.+ 

5.  What  is  the  value  of  ,25  of  an  hour  ? 

Ans.  15  minutes. 

6.  What  is  the  value  of  ,857  of  a  day  ? 

Ans.  20h.  34m.  4s.  + 

7.  What  is  the  value  of  ,125  of  a  gallon  ? 

Ans.  1  pint. 

CASE  V.     . 
To  find  the  value  of  any  decimal  of  a  pound  by  inspection. 

RULE.  —  Double  the  first  figure  or  place  of  lOths  for 
shillings,  and  if  the  second  be  five,  or  more  than  5,  add 
another  shilling,  then  call  the  figures  in  the  second  and 
third  places,  after  the  5  (if  contained)  is  deducted,  far- 
things ;  abating  1  if  their  number  is  more  than  12,  and 
two  if  more  than  36  ;  the  result  will  be  the  answer. 

EXAMPLES. 

1.  Find  the  value  of  ,785^  by  inspection. 
I4s.=double  7. 
Is.  for  5  in  the  place  of  lOths. 
8f  =35  farthings. 
i  abated  for  the  excess  above  12. 


15s.  8£d.     Answer. 


88  REDUCTION    OF    CURRENCIES. 

2.  Find  the  value  of  ,976£  by  inspection. 

Ans.  19s. 

3.  Find  the  value  of  ,542<£  by  inspection. 

Ans.  10s.  lOd. 

4.  Find  by  inspection  and  add  together  the  values  of 

,  ,875^,  ,096£,  ,763^,  and  ,008£. 

Ans.  £1  19s. 


REDUCTION  OF  CURRENCIES. 

CASE  I. 

To  reduce  the  currencies  of  the  several  United  States,  where 
a  dollar  is  an  even  number  of  shillings,  to  Federal  Money. 

RULE  1.  —  When  the  sum  consists  of  pounds  only,  an- 
nex a  cipher  to  it,  and  divide  by  half  the  number  of  shil- 
lings in  a  dollar  ;  and  the  quotient  will  be  dollars  ;  if 
there  be  a  remainder,  annex  ciphers  to  it,  and  divide 
again,  by  which  you  will  get  the  cents,  &c. 

2.  But  if  the  sum  consists  of  pounds,  shillings,  pence, 
&c.  bring  it  into  shillings,  and  reduce  the  pence  and  far- 
things to  a  decimal  of  a  shilling  ;  annex  said  decimal  to 
the  shillings,  with  a  decimal  point  between  ;  then  divide 
the  whole  by  the  number  of  shillings  contained  in  a  dol- 
lar, and  the  quotient  will  be  dollars,  cents,  mills,  &c. 

EXAMPLES. 

1.  Reduce  ,£73  New-England,  Virginia,  &c.  currency, 
to  Federal  Money. 

3)730 

$243,33£cts.     Ans. 

2.  Reduce  .£45  15s.  7^d.  New-England  currency  to 
Federal  Money. 

20 

s.  --  d.  £d.=,5 

A  dollar  =6)915,  625  12)7,500 

-  ,  -  .  -  d. 

$152,  604+  Ans.       ,625  decimal  =7£ 
c.  m. 


REDUCTION    OF    CURRENCIES.  89 

3.  Reduce  ^105 14s.  3fd.  New-York  and  North-Car-    ' 
olina  currency  to  Federal  Money. 

£.       s.       d. 

105     14     3f  f =,75  of  a  penny. 

20  d. 

s. 12)3,7500 

Adollar=8)2114,  3125  

• ,3125  decimal  =f 

$264,  2S9T°o6<y+  Ans. 
c.  m. 

4.  Reduce  <£431    New-York    currency    to    Federal 
Money. 

4)4310 

$1077,  50cft>.  Ans. 

RULE  2. — Bring  the  given  sum  into  a  decimal  expres- 
sion by  inspection,  as  in  case  3  of  decimal  fractions  ; 
then  divide  the  whole  by  ,3  in  New-England,  &c.  cur- 
rency, and  by  ,4  in  New-York,  &c.  currency  ;  and  the 
quotient  will  be  dollars,  cents,  &c. 


EXAMPLES. 


1.  Reduce   <£54   8s.   5±d.  New-England  currency    to 
Federal  Money.  £. 

,3)54,415  decimally  expressed. 

$18l,3Sc.+  Ans. 

2.  Reduce  75.   life?.  New-England   currency  to   Fed- 
eral Money. 

7s.  ll£r/.=<£0,399  decimally  expressed. 
Then  ,3),399 

$l,33c.  Ans. 

3.  Reduce  <£513  16s.  lOt/.  New-York,   &c.   currency 
to  Federal  Money. 

£. 
,4)513,842  decimally  expressed. 

$J284,60£c.  Ans. 
H3 


90  REDUCTION  OF  CURRENCIES. 

4.  Reduce  19s.  5£d.  New-York,  &c.  currency  to  Fed- 
eral Money.  <£. 

,4)0,974  decimal  of  19s.  5£d. 

$2,  43£c.  Ans. 

NOTE. — By  the  preceding  rule,  you  may  carry  on  the 
decimal  to  any  degree  of  exactness  ;  but  in  ordinary  prac- 
tice, the  following  Contraction  may  perhaps  be  useful. 

RULE  3. — To  the  shillings  contained  in  the  given  sum, 
annex  8  times  the  given  pence,  increasing  the  product  by 
2 ;  then  divide  the  whole  by  the  number  of  shillings  con- 
tained in  a  dollar,  and  the  quotient  will  be  cents. 

EXAMPLES. 

1.  Reduce  45s.  6d.  New-England,  &c.    currency,  to 
Federal  Money.  

s.  6x8+2=50  to  be  annexed. 

6)45,50 

$7,58f  c.  Ans.  or  6)4550 

$c. 

758+ cents  =7,58 

2.  Reduce  £2  10s.  9d.  New-York,  &c.  currency,  to 
Federal  Money.  £2  I0s.=50s, 


9x8+2=74  to  be  annexed. 
Then  8)5074  s. 

$  c.        Or  thus  8)50,74 

634+cents=6,34 

$6,34|c.  Ans. 

NOTE. — When  there  are  no  pence  given  in  the  sum, 
you  must  annex  two  ciphers  to  the  shillings  ;  then  divide 
as  before,  &c. 

3.  Reduce  <£3  5s.  New-England  currency,  to  Federal 
Money. 

«£35s.=65s.    Then  6)6500 

$  c. 

1083£cts.=  10,83£  Ans. 


REDUCTION  OF  CURRENCIES.  91 

CASE?  II. 

To  reduce  the  currency  of  New-  Jersey,  Pennsylvania, 
Delaware,  and  Maryland,  to  Federal  Money. 

Rule.  —  Multiply  the  given  sum  by  8,  and  divide  the 
product  by  3,  and  the  quotient  will  be  dollars,  &c. 

EXAMPLES. 

1.  Reduce  <£245  New-Jersey,  &c.  currency,  to  Fed- 
eral Money. 

f  653,33£c.  Ans, 


NOTE.  —  When  there  are  shillings,  pence,  &c.  in  the 
given  sum,  reduce  them  to  the  decimal  of  a  pound,  then 
multiply  and  divide  as  in  the  preceding  question. 

2.  Reduce  =£36  Us.  8^d.  New-Jersey,  &c.  currency, 
to  Federal  Money. 

j£36,5354  decimal  value. 

8 

3)292,6832 

$97,561-^+  Ans. 
c.  m. 

3.  Reduce  17s,  9d.  New-Jersey,  &c.  to  Federal  Mon- 
ey. Ans.  $2,36c.  6f  m. 

CASE  III. 

To  reduce  the  currency  of  South-  Carolina,  and  Georgia, 
to  Federal  Money. 

RULE.  —  Multiply  the  given  sum  by  30,  and  divide  the 
product  by  7  ;  the  quotient  will  be  dollars,  cents,  &c. 
Or,  multiply  by  3  and  divide  by  ,7. 

EXAMPLES. 

1.  Reduce  <£100  South-Carolina  and  Georgia,  curren- 
cy, to  Federal  Money. 

^100x30=3000;  3000--7  =$428,57  14+  Ans. 


92  REDUCTION  OF  CURRENCIES. 


2.  Reduce  £5±  J  6s.  9f  £  Georgia  currency  to  Feder- 
al Money. 

54,8406  decimal  expression, 
30 


7)1645,2180 

$235,0311+  Ans. 

3.  Reduce  Us.  6d.  South-Carolina,  &c.  to  Federal 
Money.  Ans.  $2  46c.  4m.  + 

CASE  IV. 

To  reduce  the  currency  of  Canada  and  Nova-Scotia,  to 
Federal  Money. 

RULE. — Multiply  the  given  sum  by  4,  if  it  contain 
pounds  only,  and  the  product  will  be  dollars.  If  it  con- 
tain shillings,  reduce  the  whole  to  shillings,  and  divide  by 
5  ;  if  it  contain  pence,  reduce  the  whole  to  pence,  and 
divide  by  60  ;  and  the  quotient,  in  either  case,  will  be  dol- 
lars ;  to  the  remainders,  if  there  be  any,  annex  ciphers, 
and  continue  the  division,  by  which  you  will  obtain  the 
cents  and  mills. 

Or,  when  the  given  sum  contains  shillings,  pence,  &c. 
reduce  them  to  the  decimal  of  a  pound,  annex  the  deci- 
mal to  the  pounds,  and  multiply  the  whole  by  4  ;  the  pro- 
duct will  be  dollars,  cents,  &c. 

EXAMPLES. 

1.  Reduce  ^6125  Canada  and  Nova-Scotia  currency,  to 
Federal  Money. 

125 
4 


$500  Ans. 

2.  Reduce  ^68  14s.  Nova-Scotia  currency,  to  Federal 
Money. 

£.  s.  £.    , 

68  14  Or  68,  7  decimal  expression, 

20  4 


5)  1 374  $274,8  dimes.  Ans. 

f  274,80cts.  Ans. 


DEDUCTION    OF    CURRENCIES.  93 

3.  Reduce  £45  17s.  9d.  Canada  and  Nova-Scotia  cur- 
rency to  Federal  Money. 

£.  s.  d. 

45  179  £. 

20  Or  45,8875  decimal  expression. 

4 

917  

12  $183,5500 

6,0)1101,3 

$183,55cts.  Ans. 

4.  Reduce  £58   13s.   6£d.   Canada,   &c.   to  Federal 
Money. 

£.     s.     d. 

58  13    6£    £=,5  of  a  penny. 
20  £. 

—  Or  58,67708  decimal  expression. 

1173  4 

12  $.     c.m. 

8234,70832=234,708T3<&+Ans. 
6,0)1403,2^ 

$234,708£  Ans. 
c.  m. 

5.  Reduce  .£528    17s.  8d.    Canada,  &c.  to  Federal 
Money.  Ans.  $2115,53cts.+ 

6.  Reduce  £1   2s.   6d.  Nova-Scotia,  &c.    money,  to 
Federal  Money.  Ans.  $4,50cts. 

7.  Reduce  13s.    ll^d.   Nova-Scotia,   &c.   money,  to 
Federal  Money.  Ans.  $2,79cts.+ 

CASE  V. 

To  reduce  the  money  of  Great  Britain  to  Federal  Money. 

RULE. — If  the  given  sum  be  pounds  only,  multiply  by 
40,  and  divide  by  9,  or  multiply  by  4,  and  divide  by  ,9 ; 
the  quotient  will  be  dollars ;  if  there  be  any  remainder, 
annex  ciphers  to  it,  and  continue  the  division  ;  the  quo- 
tient will  be  cents,  &c.  But  if  it  consist  of  pounds  and 
shillings,  reduce  it  to  shillings,  then  double  them,  and 
divide  as  before.  And  if  it  contain  pounds,  shillings, 


94  REDUCTION  OF  CURRENCIES. 

and  pence,  reduce  it  to  pence,  and  divide  by  54,  the  num- 
ber of  pence  in  a  dollar. 

Or,  when  the  sum  consists  of  pounds,  shillings,  and 
pence,  reduce  the  shillings,  &/c.to  the  decimal  of  a  pound, 
then  multiply  the  whole  by  '40,  and  divide  by  9. 

EXAMPLES. 

1.  Reduce  £36  sterling  into  Federal  Money. 

36x40 

=$160  Ans. 

9 

2.  Reduce  ^636  9s.  sterling  into  Federal  Money. 

£.  s.  £. 

36  9  Or  36,45  decimal  expression. 

20  40 

729  9)1458,00 

o  

$162  Ans. 

9)1458  doubled. 


$162  Ans. 
3.  Reduce  ,£579  17s.  9d.  sterling  into  Federal  Money. 

£.  s.  d. 
579  17  9  £. 

20  Or  579,8875  decimal  expression, 
40 


11597 


12  9)23195,5000 

c.  m. 


54)139173(2577,277+  An.$2577,2777+Ans. 

4.  Reduce  <£100  sterling  money,  to  Federal  Money. 

Ans.  $444,44|c. 

To  reduce  Federal  Money  to  the  currency  of 

Neu^England,  j         Multiply  the  given  sum  by  ,3 
Virginia,       I  J  J  &  the  product  will  be  pounds, 


Kentucky,  and  f*  \  ^  jg^to  of  a  pound. 

Tennessee.      ) 

„  „  7  ,  .  .  (  Multiply  the  given  sum  by  ,4 
£"*?%*  FA  H  |  and  the  product  will  be  pounds, 
North-Carohna.  5«  \  and  decimals  of  a  pound. 


REDUCTION    OF    CURRENCIES. 


96 


3. 


4. 


New-Jersey, 
Pennsylvania, 
Delaware  and 

Maryland. 

South-Carolina  )  j> 
and  Georgia.  ]  & 


Multiply  the  given  sum  by  3, 
and  divide  the  product  by  8, 
and  the  quotient  will  be  pounds 
and  decimals  of  a  pound. 

Multiply  the  given  sum  by  ,7 
and  divide  the  product  by  3, 
and  the  quotient  will  be  pounds 
and  decimals  of  a  pound. 


EXAMPLES  IJV  THE  FOREGOING  RULES. 

1.  Reduce  $152,6Qcts.  to  New-England  currency. 
152,00 
,3 

.£45,780  Ans.  ,£45  15s.  7d.  TV 

20  But  the  value  of  any  decimal  of  a  pound, 

may  be  found  by  inspection,   as  in  case   5, 

s.  15,600  of  Decimal  Fractions,  page  87. 
12 


d.  7,200 

2.  Reduce  $198  into  Virginia,  &c.  currency. 
196 
,3 

£58,8  Ans.  =^58  16s. 

3.  In  $629,  how  many  pounds  New-  York,  &c.  currency  ? 
629 
,4 


.=^251  12s. 

4.  Bring  $I10,51cts.  into  New-Jersey,  &c.  currency. 
110,51 

3     Double  ,4  -8s.     Then  take  2  from  41  = 
-     39  for  fart  lungs  =9d.  3qrs.    See   Case  5 
8)331,53     of  Decimals,  page  87. 

<£41,441  Ans.  =£41  8s.  9fd.  by  inspection. 


96  KEDUCTION  OF  CURRENCIES. 

5.  How  many  pounds,  &c.  South-Carolina,  &c.  cur- 
rency in  $65,36cts.  Smills  ? 
65,368 

,7 

3)45,7576 


^15,25253  Ans.=,£15  5s. 


CASE  II. 

To  reduce  Federal  Money  to  Canada  and  Nova-Scotia 
currency. 

RULE.  —  Divide  the  dollars  by  4,  and  the  quotient  will 
be  pounds  ;  to  the  remainder,  if  there  be  any,  annex  the 
cents,  &c.  and  to  that  number,  annex  a  cipher  ;  then  halve 
that  number,  and  cut  off  the  left  hand  figure  or  figures 
less  than  20,  for  shillings  ;  the  remaining  figure  or  figures 
multiply  by  12,  and  cut  off  just  as  many  right  hand  fig- 
ures from  the  product  as  you  multiply  ;  the  left  hand 
ones  are  pence,  &c. 

Or,  multiply  the  given  sum  by  60,  the  number  of  pence 
in  a  dollar,  and  if  it  contains  cents,  cut  off  two  figures  on 
the  right,  if  mills,  three  ;  those  on  the  left  are  pence, 
which  must  be  reduced  into  pounds.  The  figures  cut  off 
will  be  decimals  of  a  penny. 

Or,  lastly,  divide  the  given  sum  by  4,  and  the  quotient 
will  be  pounds,  and  decimals  of  a  pound. 

EXAMPLES. 

1.  Reduce  $183,55cts.  into  Canada  and  Nova-Scotia 
Money. 

$ 
4)183 

,£45  —  and  3  remain  to  be  placed  before  the 
55  cents.  355,  to  which  annex  a  cipher.  3550  halved, 
or  divided  by  2=1775,  of  which  cut  off  the  two  left  hand 
figures,  as  they  are  less  than  20,  which  are  shillings  17,75. 
75x12=900,  of  which  cut  off  the  two  right  hand  figures 
because  you  multiply  two  =9,00;  and  the  9  on  the  left 
are  pence  ;  the  answer  is,  therefore,  ^45  17s.  9d. 


REDUCTION    OF    CURRENCIES.  J)7 

By  the  second  method :  183,55x60=1101300,  from 
which  cut  off  the  two  right  hand  figures,  and  it  is  11013, 
which  are  pence.  These  reduced,  are  £45  17s.  9d.  Ans. 

By  the  last  method  :         $.  c. 
4)183,55 

£45,8875  =£45  17s.  9d.  Ans. 

2.  Bring  S741  into  Nova-Scotia  currency. 

Aris.  £185  5s. 

3.  What  sum,  Nova-Scotia  money,  is   equal  to   $311, 
75cts.  ?  Ans.  ,£77  18s.  9d. 

4.  In  $2907,56cts.,  how  much  Nova-Scotia  money  1 

Ans.  £726  l?s.  9£d.+ 

5.  How  many  pounds,  &c.   Canada  money  are    in 
$2114,50cts.  ?  Ans.  ,£528  12s.  6d. 

CASE  III. 

To  reduce  Federal  Money  to  the  money  of  Great  Britain. 

RULE. — Multiply  the  given  sum  by  9,  and  divide  the 
product  by  40 ;  or  multiply  the  given  sum  by  ,9  and  di- 
vide the  product  by  4  ;  and,  in  either  case,  the  quotient 
will  be  the  answer  in  pounds,  and  decimals  of  a  pound. 

EXAMPLES. 

1.  Reduce  $l83,55cts.  into  sterling  money. 
183,55 
9 

Or  183,55 

4,0)165,195  ,9 

£41,29875        4)165,195 
20 


£41,29875=^41    5s.   ll£d.  by  in- 


s.5,97500  spection, 

12 

d.  11,700 
4 


s.  lid.  2,8qrs.  Ans. 

2.  Bring  $247,44c.  5m.  into  English  money. 

Ans.  £55  13s. 

3,  Show  the  value  of  $1000  in  British  money. 

Ans.  £225. 


98  RULE    OF    THREE. 

4.  Tell  me  what  sum,  in  sterling  money,  is  just  equal 
to  $2466,33cts.  3£mills.  Ans.  ^554  18s.  6d. 


RULE  OF  THREE. 

THE  RULE  OF  THREE  teaches  to  find  a  number  having 
the  same  proportion  to  a  given  number,  that  two  other 
given  numbers  have  between  themselves.  For  this  reason 
it  is  sometimes  called  the  Rule  of  Proportion.  It  is  call- 
ed the  Rule  of  Three,  because  in  each  of  its  questions 
there  are  given  three  numbers  at  least.  And  because  of 
its  excellent  and  extensive  use,  it  has  been  often  named 
the  Golden  Rule. 

RULE.  —  Write  down  the  number,  which  is  of  the  same 
kind  with  the  answer  or  number  required. 

Consider  whether  the  answer  ought  to  be  greater  or  less 
than  this  number  ;  if  greater,  write  the  greater  of  the 
two  remaining  numbers  on  the  right  hand  of  it  for  the 
third,  and  the  other  on  the  left  for  the  first  number  or 
term  ;  but  if  it  ought  to  be  less,  write  the  less  of  the  two 
remaining  numbers  in  the  third  place,  and  the  other  in 
the  first. 

Multiply  the  second  and  third  terms  together,  divide  the 
product  by  the  first,  and  the  quotient  will  be  the  answer. 

NOTE  1.  —  It  is  sometimes  most  convenient  to  multiply 
and  divide  as  in  Compound  Multiplication  and  Division. 
But  when  it  is  not,  then  reduce  each  of  the  compound 
terms  to  the  lowest  denomination  mentioned  in  it,  and 
reduce  the  first  and  third  to  the  same  denomination  ;  then 
will  the  answer  be  of  the  same  denomination  with  the 
second  term.  And  the  answer  may  afterwards  be  brought 
to  any  denomination  required. 

2.  When  there  happens  to  be  a  remainder  after  the 
division,  reduce  it  to  the  name  next  below  the  last  quo- 
tient, and  divide  by  the  same  divisor  ;  so  shall  the  quo- 
tient be  so  many  of  the  said  next  denomination  ;  do  this 
as  long  as  there  is  any  remainder,  or  till  you  have  reduc- 
ed it  to  the  least  name,  and  all  the  quotients  together 
will  be  the  answer. 


RULE    OF    THREE.  99 

3.  If  the  first  term,  arid  either  the  2d  or  3d,  can  be  di- 
vided by  any  number,  without  remainder,  let  them  be  di- 
vided, and  the  quotients  used  instead  of  them. 

4.  There  are  four  other  methods  of  operation   besides 
the  general  one  above  delivered,   any  of   which,  when 
possible,  performs  the  work  much  shorter  than  it.     They 
are  thus  : 

First,  Divide  the  2d  term  by  the  1st,  multiply  the  quo- 
tient by  the  3d,  and  the  product  will  be  the  answer. 

Second,  Divide  the  3d  term  by  the  1st,  multiply  the 
quotient  by  the  2d,  and  the  product  will  be  the  answer. 

Third,  Divide  the  1st  term  by  the  2d,  divide  the  3d  by 
the  quotient,  and  the  last  quotient  will  be  the  answer. 

Fourth,  Divide  the  1st  term  by  the  3d,  divide  the  2d 
by  the  quotient,  and  the  last  quotient  will  be  the  answer. 

Two  or  more  statings  are  sometimes  necessary,  which 
may  always  be  known  from  the  nature  of  the  question. 

The  method  of  proof  is  by  inverting  the  question. 

EXAMPLES. 

1.  What  is  the  value  of  Zjfo.  6oz.  I9dwt.  of  gold,  at 
^3  19s.  lid.  an  ounce? 

oz.  •£.  s.  d.  }£>.  oz.  dwt. 

1       :       3  19  11     :  :         2    6    19 
20  20  12 

20  79  30 

12  20 

959  619 

619 

8631 
959 
5754 

2,0)59362,1 

12)2968 1 ^    Pence,  which  divided  by   12  and 
20,  gives  the  answer  in  pounds, 
2,0)247,3s.  5d.     &c. 

Ans.  £123  13s. 


100  KULE    OF    THREE. 

2.  If  9}fc.  of  tobacco  cost  I  dollar  20  cents,  what  will 
.  cost  ? 

ft-  8-  cts.  ft. 

9       :         1,20      :  :       25 
120 

9)3000 

§3,33£  Ans. 

3.  If  25ft.  of  tobacco  cost  $3,33£,  what  will  Oft.  cost  ? 
ft-  *•  cts.  ft. 

25       :        3,33i     :  :     9 
9 

25)3000(120  cents,  or  $1,20  the  An. 
25 

50 
50 

0 

4.  What  is  the  value  of  a  firkin  of  butter  containing 
56ft.  at  I0£d.  per  pound  ? 

ft. 
:        56 


<£2  9  0  0  the  Answer. 

5.  If  7cwt.  Iqr.  of  sugar  cost  36dols.    10c.,  what  will 
be  the  price  of  43cwt.  2qr,  ?  Ans.  g216,  60c. 

cwt.gr. 
:      43    2 
4 

174 


29)628140(21660cts.=$216,  60cts. 


cwt.  qr. 

$.  cts. 

7  I 

:   36,10 

4 

17  4 

29 

1444 

2527 

361 

RULE    OF    THREE.  101 

6.  If  6  horses  eat  21  bushels  of  oats  in  a  month,  haw 
many  bushels  will  20  horses  eaf  in  the  same  time.? 

Hor.  Bus.  Hor.  Bus. 

6  :         21     ::       20       :       70     Ans. 

7.  A  man  bought  sheep  Idol,  llcts  per  head,  to  the 
amount  of  51dol.  6cts  :  how  many  sheep  did  he  buy  1 

$  cts.         sh.  $  cts.  sh, 

1,11     :       1     ::     51,06       :     46  Ans. 

8.  What  is  the  value  of  an  cwt.  of  sugar  at  5^d.  per 
ft.,  25ft.  a  qr.  ? 

ft-  d.     qr.  ft.  &s.d- 

1          :         52         :  :     100     :        2  5  10  Ans. 

9.  How  much  in  length  of  that  which  is  4J  inches 
broad  will  make  a  square  foot  ? 

Breadth.         Length.         Breadth.         Length. 
4£         :         12       :':         12       :      2ft.  Sin.  Ans. 

10.  Bought  6  casks  of  raisins,  each  weighing  Icwt.  lq. 
12^1b.  ;  what  will  they  come  to  at  £2  Is.  fed.  per  cwt.  ? 

Cwt.  £  s.  d.  Cwt.  qr.   fb-  £   s.  d. 

1:    2     18       ::        1        I      12^  +  6  :  17  0  4£+  Ans. 

11.  If  a  man  spend  2  dollars  45  cents  a  week,   what 
will  it  amount  to  in  a  year  ? 

d'ys.  $.  cts.          d'ys.  $.  cts. 

7  :         2,45     :  :     365     :     127,75  Ans. 

12.  What  is  the  value  of  a  pipe  of  wine  at  10-J-d-.  per 
pint  ? 

Pint.          d.  Pipe.  £.     s. 

1       :       10£     ::       1         :  44    2  Ans. 

13.  How  many  quarters  of  corn  can  I  buy  for  280  dol- 
lars, at  f  of  a  dollar  per  bushel  ? 

Ans.  52  quarters,  4  bushels. 

14.  What  is  the  value  of  2qrs.  Ina.  velvet,  at  19s.  8^d. 
per  ell  English  ?  Ans.  8s.  lO^d.  TV 

15.  Suppose  18  yards  of  broadcloth  l^yds.  wide  is   to 
be  lined  with  shalloon  that  is  f  of  a  yard  wide  ;  how  many 
yards  of  shalloon  will  be  sufficient  !  Ans.  36  yds. 

16.  If  52  yards  of  cloth  cost  156  dollars,    how   much 
will  4  yards  cost  ?  Ans.  12  dollars. 

17.  Bought  36  yards  of  cloth  for  108  dollars,  and  sold 
the  same  at  3|  dollars  per  yard ;  how  much  did  I  gain  ? 

Ans.  18  dollars. 
I  2 


«ULE    OF    THREE. 

18.  If  7yds.  of  ribbon  cost  3s.  4d.',  what  will   126yds. 
cost  ?  Ans.  £3. 

19.  If  a  man  earn  64*dollars  in  4  months,  how  long 
must  he  work  at  the  sanie  rate  to  pay  a  debt  of  300  dol- 
lars ?  Ans.  18  months,  3  weeks. 

20.  If  an  ounce  of  silver  be  worth  1  dollar,  10  cents, 
what  is  the  value  of  10  silver  spoons,  each  weighing  loz. 
4  pennyweights  ?  Ans.  13dolls.  20cts. 

21.  If  8f  yards  cost  4  dollars  20  cents,  what  will  13£ 
yards  cost  I  Ans.  Gdolls,  48cts. 

22.  How  long  will  it  take  5  men  to  do  the  same  work 
which  37  men  can  do  in  15  days  1  Ans.  Ill  days. 

23.  what  will  4  hogsheads  of  wine  come  to  contain- 
ing, viz.  79|,  84^,  101J,  and  112  gallons,  at  6s.  9d.  per 
gallon  ?  Ans.  £127  4s.  9d. 

24.  Bought  3hhds.  of  sugar,  each  weighing  8cwt.  Iqr. 
12fty  at  7dolls.  26cts.  per  cwt ;  what  come  they  to,  25fc. 
to  the  qr  ?  Ans.  $l82,29c.  8f  m. 

25.  If  a  chest  of  hyson  tea  weighing  79{fe.    neat,  cost 
£32  lls.  9d.,  what  is  it  per  pound  1  Ans.  8s.  3d. 

26.  B  owes  £2119  17s.  6d.,  and  he  is  worth  but  ,£1324 
18s.  5£d. ;  if  he  delivers  this  to  his  creditors,  how  much 
do  they  receive  on  a  pound  ?  Ans.  12s.  6d. 

%27.  A  merchant  failing  in  trade,  owes  in  all  29475  dol- 
lars, and  delivers  up  his  whole  property  worth  21894  dol- 
lars, 3  cents  ;  how  much  per  cent,  does  he  pay  ;  and  what 
is  B's  loss  to  whom  he  owed  325  dollars  ? 

Ans. — He  pays  $74,28cts.  per  cent, 
and  B  loses  f  83,59cts. 

28.  If  a  staff,  4  feet,  8  inches  in  length,  cast  a  shadow 
6  feet ;  how  high  is  that  steeple  whose  shadow  is  153  feet  ? 

Ans.  119  feet. 

29.  Bought  270  quintals  codfish  for  780  dollars  ;  freight 
37  dollars,  70cts. ;  wharfage,  truckage  and  other  expenses 
30  dollars,  60cts.  ;  at  what  must  I  sell  them  per  quintal, 
so  as  to  gain  143  dollars  on  the  whole  ? 

Ans.  £3,67cts.  1m. + 

30  If  f  of  a   farm   cost   $1081,   what   is   the    whole 
worth  1 

fif.       $.  fif.         $.     cts.  m. 

3  :  1081    :  :    5    :    1801,66  6+Ans. 


RULE    OF    THREE-  103 

31.  If  a  man  spend  46  cents  a  day,  what  will  it  amount 
to  in  a  year  ?  Ans.  $  1 67,90. 

32.  Lent  a  friend  292  dollars  for  six  months  ;  sometime 
afterward  he  lent  me  403  dollars  ;  how  long  must  I  keep 
it  to  balance  the  favour  ?  Ans.  4m.  lw^2d.+ 

33.  If  100  dollars  gain  6  dollars  interest  in  one  year, 
how  much  will  480  dollars  gain  in  the  same  time  ? 

Ans.  828,80. 

34.  If  480  dollars,  gain  28  dollars,  80  cents  in  one  year, 
>  how  much  will  it  gain  in  87  days  ? 

Ans.  $6,86cts.  4m.  + 

35.  How  much  land,  at,  2  dollars,  50  cents  per  acre, 
must  be  given  in  exchange  for  360  acres,  at  3  dollars,  75 
cents  per  acre  ?  Ans.  540  acres. 

36.  Bought  a  silver  cup  weighing  9  ounces  4  penny- 
weights 16  grains,  for  £3  2s.  3d.  3qrs.f ;  what  was  it  per 
ounce  ?  Ans.  6s.  9d. 

37.  There  is  a  cistern  which  has  four  cocks ;  the  first 
will  empty  it  in  10  minutes,  the  second  in  20  minutes, 
the  third  in  40  minutes,  the   fourth   in   80  minutes;  in 
what  time  will  all  four  running  together  empty  it  ? 

Ans.  5min.  20sec. 

38.  A  hired  two  men,,B  and  C,  to  cut  wood  for  50cts. 
per  cord  ;  B  could  cut  a  cord  in  4  hours,  C  in  6  hours ; 
how  long  would  it  take  both  to  cut  1  cord  ? 

Ans.  2  hours,  24  minutes. 

39.  If,  when  wheat  is  6s.   3d.  per  bushel,  the  penny 
loaf  weigh  9  ounces,  what  ought  it  to  weigh  when  wheat 
is  8s.  2£d.  per  bushel  ?  Ans.  6oz.  13drs.+ 

40.  When  a  man's  yearly  income  is  949   dollars,  how 
much  is  it  per  day  ?  Ans.  $2,60cts. 

41.  What  is  the  commission  on  1525  dollars  at  4|dolls. 
per  cent.  ?  Ans.  g6S,62cts.  5m. 

42.  What  will  374  feet  of  boards  come  to  at  l^cerits 
per  foot!  Ans.  $5,61  cts. 

43.  What  will  39  thousand  6  hundred  and  30  casts  of 
staves  come  to  at  15  dollars  50  cents  per  thousand  ? 

Ans.  $6l4,73cts. 

NOTE. — 2  staves  make  1  cast;  50  casts  1  hundred; 
10  hundred  1  thousand  ;  in  Maine,  by  a  late  law  of  this 
State. 


104  RULE    OF    THREE. 

44.  If  the  inventory  of  a  town  be  358400  dollars,  upon 
which  there  is  assessed  a  tax  of  850  dollars,  what  will  it 
be  on  a  dollar  ;  and  what  will  B's  tax  be,  whose   estate 
in  that  town  is  valued  at  1792  dollars  ? 

„   (  2mills  T3o^+  on  a  dollar,  and 
s§  \      B's  tax  will  be  g4,25cts. 

45.  What  will  the  charter  of  a  ship  of  306  tons  amount 
to,  from  May  28  to  October  10th  following,  at  2  dollars 
per  ton,  per  month  of  30  days  1  . 

Ans.  2774  dollars,  40  cents. 

NOTE. — The  days  of  receiving  and  discharging  are 
both  included. 

46.  If  4^  hundred  weight  may  be  carried  36  miles  for 
35s.  how  many  pounds  can  I  have  carried  20  miles  for  the 
same  money,  25}fo.  a  quarter.  Ans.  810ffc. 

47.  Sold  a  ship  for  £537,  and  I  owned  f  of  her ;  what 
was  my  part  of  the  money  1  Ans.  <£201  7s.  6d. 

48.  What  quantity  of  water  must  I  add   to   a  pipe   of 
mountain  wine  valued  at  ,£33,  to  reduce  the  first  cost  to 
4s.  63.  per  gallon  1  Ans.  20f  gallons. 

49.  A  and  B  depart  from  the  same   place,  and   travel 
the  same  road  ;  but  A  goes  6  days  before  B,  at  the  rate  of 
21  miles  a  day  ;  B  follows  at  the  rate  of  28  miles  a  day  ; 
in  what  time  and  distance  will  he  overtake  A  1 

.        MS  days. 
3*  \  504  miles. 

50.  A  factor  bought  a  certain  quantity  of  broadcloth 
and  drugget,  which,  together,  cost  £81  ;  the  quantity  of 
broadcloth  was  50  yards,  at  18s.  per  yard,  and  for  every 
5 yards  of  broadcloth  he  had  9  yards  of  drugget ;  I  de- 
mand how  many  yards  of  drugget  he   had,   arid   what  it 
cost  him  per  yard  !  Ans.  90yds.  at  8s.  per  yd. 

51.  If  60  gallons  of  water,  in  one  hour,  fall  into  a  cis- 
tern containing  300  gallons,  and  by  a  pipe  in  the  cistern, 
35  gallons  run  out  in  forty  minutes  ;  in  what  time  will  it 
be  filled  1  Ans.  40  hours. 

52.  How  many  yards  of  .cloth  3qrs.  wide,  will  be  equal 
in  measure  to  30  yards,  5qrs.  wide.  Ans.  50yds. 

53.  Bought  a  pipe  of  wine  for  84  dollars,  and  found  it 


PRACTICE.  105 

leaked  out  17  gallons;  I  sold  the  remainder  at  12£cts.  a 
pint  ;  did  I  gain  or  lose,  and  how  much  ? 

Ans.  I  gained  30  dols. 

54.  How  many  yards  of  paper,  3  quarters  wide,  will 
paper  a  room,  30  feet  long,  24  wide,  and  12  high,  deduct- 
ing 81  square  feet  for  fire-place,  door,  and  windows  1 

Ans.  180  yards. 

55.  A  garrison  consisting  of  1500  men,  being  besieg- 
ed, have  provisions  for  three  months  only  ;  but  it  being 
necessary  they  should  hold  out  five  months,  how  many 
men  must  depart,  that  the  same  provisions  may  serve  that 
time  ?  Ans,  600  men. 

56.  A  regiment  of  soldiers  consisting  of  1000  men,  are 
to  have  new  coats,  and  each  coat  is  to  contain  2yds.  and 
Iqr.  of  cloth  that  is  5  quarters  wide  ;  how  much  shalloon, 
that  is  3  quarters  wide,  will  line  them  1     Ans.  3750yds. 

57.  A  merchant  shipped  for  the   West-Indies  39000 
feet  of  boards,  at  $8,20cts.  per  M.  ;  300  quintals  offish, 
at  $2,60cts.  per  quint.  ;  15000  shingles,  at  $2,20cts.  per 
M.  ;  34000  hoops,  at  gl,60cts  per  M.  ;    and  glOOO  in 
cash;  and  in  return,  he  had  3000   ffo.   of  indigo  at  56 
cents  per   jfo.  2580  gallons  of  molasses,   at  20cts.  per 
gal.  ;  1000  pounds  of  coffee,  at  18cts.  per  jfc.  ;  and  18cwt. 
of  sugar,  at  $4,50cts.  per  cwt.  ;  and  his  charges  on  the 
voyage  were  $l53,80cts.     Did  he  gain,  or  lose,  by  this 
voyage?  Ans.  He  gained  SI  16. 


PRACTICE. 

PRACTICE  is  a  contraction  of  the  Rule  of  Three,  when 
the  first  term  happens  to  be  a  unit,  or  1  ;  and  is  a  concise 
method  of  ascertaining  the  value  of  goods,  &,c.  where 
money  is  reckoned  in  pounds,  shillings,  and  pence  ;  but, 
since  reckoning  in  Federal  Money,  in  all  kinds  of  busi- 
ness, has  become  universal  in  our  country,  it  has  here 
grown  into  almost  total  disuse.  I  shall,  therefore,  pre- 
sent few  examples  in  this  Rule  to  the  attention  of  the 
student  ;  arid  these  chiefly  in  Federal  Money.  A  table 
of  aliquot  or  even  parts  of  weight  he  will  find  annexed  to 
Case  2,  in  Tare  and  Tret,  page  109. 


10G 


PRACTICE. 


CASE  I. 

When  the  price  is  an  even  part  of  a  pound. 
RULE. — Find  the  value  of  the   given  quantity,   at  one 
pound  per  yard,  <fcc.  and  divide  it  by  that  even' part,  and 
the  quotient  will  be  the  answer  in  pounds. 

EXAMPLES. 

1.  What  will  129£  yards  cost,  at  2s.  6d.  per  yard  ? 
The  quantity  itself  is  the  price  at  £l  per  yd. 
2s.  6d.=£of  £. 

£.     s. 
2s.  6d.(£)  129  10  value  at  £1  per  yd. 


Ans.  c£16  3s.  9d.  value  at  2s.  6d.  per  yd. 

2.  What  will  461  bushels  of  oats  cost,  at   1  s.  8d.  per. 
bushel  ?          Is.  8d.r=Tu  of  a  £.  Ans.  ^38  8s.  4d. 

3.  What  will  21  1^  bus.  of  corn  cost,  at  4s  ? 

4s.  =|.  Ans.  M<2  5s. 

4.  What  will  543  bus.  of  wheat  cost,  at  6s.  8d.  7 

6s.  8d.=-J.  Ans.  ^181. 

5.  What  will  127  gallons  of  wine  cost,  at  3s.  4d.  ? 

8s.  4d.=f  Ans.  .£21  3s.  4d. 

6.  What  will  687^  bushels  of  salt  cost,  at  5s.  ? 

5s.==J.~  Ans.  ^171  17s.  6d. 

CASE  II. 

When  the  price  is  Federal  Money,  but  the  quantity  of  the 

goods  is  in  several  denominations. 

RULE.  —  Multiply  the  price  by  the  integers  in  the  given 
quantity,  and  take  parts  for  the  rest  from  the  price  of  an 
integer  ;  which  added  together,  will  be  the  answer. 


,65cts.  per 


1.  What  cost  9cwt.  iqr.  8jfe.  of  sugar  at 
cwt.  $     cts. 

lqr.]i-8,     65 
9 


lib- 


77,  85 
2,  1625 
,  5406+ 

,  0772+ 


$80,  6303+  Ans.   =$80,63cts. 


PRACTICE.  107 

2.  What  cost  7cwt.  3qr.  16ft.  of  raisins,  at  $9,58cts. 
per  cwt.  ?  Ans.  $75,61cts.  3m. + 

cwt.  qr.  ft. 

3.  12     0     7  of  Russian  iron,  at   $6,34cts.  per.  cwt., 
at  25ft.  a  qr.  ?  Ans.  $76,52cts.  3m. + 

cwt.qr.lb. 

4.  0     0  24  of  hemp,  at  $ll,91cts.  per  cwt.  1 

Ans.  $2,55cts.  2m. + 

5.  1     0  16  of  pig  lead,  at  $6,5 lets,  per  cwt.  at  25ft. 
a  qr.  ?  Ans.  $7,55cts.  If  m. 

6.  0    2  21  of  sugar,  at  $lO,24cts.,  per  cwt.,  at  25ft. 
a  qr.  ?  Ans.  f7,27^cts. 

7.  5  cords,  3ft.  Sin.  of  bark,  at  $4,50cts.  per  cord  ? 

Ans.  $24,56£cts. 

8.  3tons,  5cwt.  2qr.  of  hay,  at  $25,1  Gets  per  ton  ? 

Ans.  $82,20£cts. 

9.  1  ton,  9cwt.  3qr.  17ft.  of  hay,  at   $4,44cts.  4m. 
per  ton,  at  25ft.  a  qr.  ?  Ans.  $6,64cts.  8m. + 

10.  17jfc.  5oz.  I4dwt.  of  electary,  at  83cts.  per  fc.  ? 

Ans.  $!4,5Qcts.+ 

11.  32ac.  1  rood,  14  perches  of  land,  at  $l,l6cts.  per 
acre?  Ans.  S37,51cts.+ 

12.  I4ac.  3  roods,  5  perches  of  land,  at  $10,50cts.  per 
acre  1  Ans.  g!55,20cts.+ 

13.  6yds.  3qr.  2na.  of  cloth,  at  $7,82cts.  8m.  1 

Ans.  $53,8 Jets.  7£». 

14.  8gal.  3qts.  Ipt.  2gills  of  wine  at  $1,  96cts.  ? 

Ans.  f  17,5lfcts. 

The  Rule  may  be  applied  to  other  weights  and  meas- 
ures also.  But  perhaps  all  such  questions  may  be  better 
wrought  by  multiplication  of  decimals.  Take,  for  exam- 
ple, the  9th  question  in  this  Case.  What  will  Iton  9cwt. 
3qr.  17ft  of  hay  come  to,  at  $4,44cts.  4m.  per  ton,  at 
25ft.  a  qr.  ?  Ton.  $  $  c.  m. 

Decimal  expression  1, 496x4,444 =$6,648224 «=6,648-f 
answer. 


108  TARE    AND    TRET. 

TARE  AND  TRET. 

TARE  AND  TRET  are  practical  rules  for  deducting  cer- 
tain allowances,  which  are  made  by  merchants  and 
tradesmen  in  selling  their  goods  by  weight. 

Gross  weight  is  the  whole  weight  of  any  sort  of  goods, 
together  with  the  box,  barrel,  or  bag,  &c.  that  contains 
them. 

Tare  is  an  allowance  made  to  the  buyer  for  the  weight 
of  the  box,  barrel,  or  bag,  &c.  which  contains  the  goods 
bought,  and  is  either  at  so  much  per  box,  &c.,  at  so  much 
per  cwt.,  or  at  so  much  in  the  gross  weight. 

Tret  is  an  allowance  of  4f^.  in  every  104ffe.  for  waste, 
dust,  &c.,  or  ^  of  the  whole  tare=suttle. 

Cloff  is  an  allowance  of  2ff).  upon  every  3cwt.  or  336|fc. 

Suttle  is  the  weight  when  part  of  the  allowance  is  de- 
ducted from  the  gross. 

Neat  weight  is  what  remains  after  -all  allowances  are 
made. 

CASE  I. 

When  the  tare  is  a  certain  weight  per  box,  barrel,  or 
bag,  fyc. 

RULE. — Multiply  the  number  of  boxes,  or  barrels,  &c. 
by  the  tare,  and  subtract  the  product  from  the  gross,  and 
the  remainder  is  the  neat  weight  required. 

EXAMPLES. 

1.  In  10  casks  of  alum,  each  weighing  3cwt.  2qrs. 
.  gross,  tare  15ft,  per  cask,  how  much  neat  ? 
Cwt.     qr.     ffc. 
3        2       12 
10 


36      0        8  Gross. 
15xlO=150fb=l       1       10  Tare. 


34      2      26  neat,  the  answer. 


TARE    AND    TRET. 


109 


2.  In  241  barrels  of  figs,  each  3qrs.  lOfc.   gross,  tare 
10}k.  per  barrel,  how  many  pounds  neat  ? 

Ans.  22413ft. 

3.  What  is  the  neat  weight  of  21  hogsheads  of  tobac- 
co, each  5cwt.  2qrs.  171b.  gross,  tare  lOOlb.  per  hogshead, 
at  251b.  to  the  qr.  I  Ans.  98cwt.  71b. 

4.  What  is  the  neat  weight  of  4  chests  of  hyson  tea, 
weighing,  gross  961b.  971b.  lOllb.  and  1031b.,   tare  201b. 
per  chest  I  Ans.  3171b. 

CASE  II. 

Wlien  the  tare  is  a  certain  weight  per  cwt. 
RULE.  —  Divide  the  gross  weight  by  the  aliquot*  parts 
of  an  cwt.  contained  in  the  tare,  and  subtract  the  quotient 
from  the  gross,  and  the  remainder  is  the  neat  weight. 

EXAMPLES. 

1.  Gross  372cwt.  3qrs.  171b.,  tare  161b.  per  cwt.,  how 
much  neat  ?  Cwt.  qrs.  Ib. 

16Ib.  is  i)372    3     17 

53     1       2    t^re  subtracted. 


319     2     I4flhe  answer. 


2.  What  is  the  neat  weight  of  7  barrels  of  potash,  each 
weighing  4021b.  gross,  tare  lOlb.  per  cwt.,  at  251b.  to 
the  qr.  1  Ans.  25321b.  9|oz. 

*  An  aliquot  part  of  any  number  is  such  a  part  of  it 
as,  being  taken  a  certain  number  of  times,  exactly  makes 
that  number. — If  lOOlb.  be  taken  for  a  cwt.,  501b.=£; 


•v 


TABLE  OF  ALIQUOT  PARTS. 


Parts  of  an 

cwt. 

Parts  of  £ 

cwt. 

Parts  of 

2  qrs.  is 

i 

281b.  is 

A 

I41b.  is 

1 

i 

14 

-       i 

7 

161b.  is     - 

-  J- 

8    -      - 

1 

4     - 

14 

i 

7 

-       i 

3*      - 

8    - 

TV 

Of  501b.  25 

i 

Of  251b.  1 

7 

-       TV 

1  ^ 

J      i 

Of  100.  12^ 

10 

i 

10 

TV 

5 

TV 

5 

^V 

cwt. 


i 


R 


HO  TARE    AND    TRET. 

8.  In  I29cwt.  3qrs.  161b.  gross,  tare  141b.  per  cwt., 
what  is  the  neat  weight  ?  Ans.  113cwt.  2qrs.  17^1b. 

4.  In  25  barrels  of  figs,  each  2cwt.  Iqr.  gross,  tare 
161b.  per  cwt.,  how  much  neat  ?  Ans.  48cwt.  Oqrs.  24lb. 

CASE  III. 

When  tret  is  to  be  allowed  with  tare,. 

RULE. — Divide  the  suttle  weight  by  26,  and  the  quo- 
tient is  the  tret,  which  subtract  from  the  suttle,  and  the 
remainder  is  the  neat  weight. 

EXAMPLES. 

1.  In  9cwt.  2qrs.  171b.  gross,  tare  371b.  and  tret  as 
usual,  how  much  neat  1 


Cwt. 

qr*. 

tb- 

9 

2 

17  gross. 

1 

9  tare. 

26)9 

1 

8  tare-suttle. 

1 

12^  tret. 

8 

3 

234-4-  Answer. 

2.  In  7  casks  of  prunes,   each  weighing  4cwt.  gross, 
tare  17^1b.  per  cwt.  and  tret  as  usual,  how  much  neat,  at 
251b.  to  the  qr.  ?  Ans.  22cwt.  21-^lb. 

3.  What  is  the  neat  weight  of  3hhds.  o£  sugar  weigh- 
ing as  follows  ;  -the  first  4cwt.  51b.  gross,  tare  731b. ;  the 
second  3cwt.  2qrs.  gross,  tare  561b. ;  and  the  third  2cwt. 
3qrs.  171b.  gross,  tare  471b. ;  and  allowing  tret  to  each  as 
usual ;  251b.  a  qr.  ?  Ans.  8cwt.  Iqr.  12£lb. 

4.  What  is  the  neat  weight  of  10  casks  of  raisins,  each 
weighing  3cwt.  2qrs.,  tare  141b.  per  cwt.  tret  as  usual ; 
and  what  will  be  the  amount  at  $15  per  cwt.  ? 

Ans.  29cwt.  Iqr.  22Tyb.  $441,  70cts.  6m.-f 

CASE  IV. 

When  tare%  tret,  and  doff,  are  all  allowed. 

RULE. — Deduct  the  tare  and  tret,  as  before,  and  divide 
the  suttle  by  168,  and  the  quotient  is  the  cloff,  which 
subtract  from  the  suttle,  and  the  remainder  is  the  neat. 


DOUBLE    RULE    OF    THREE.  Ill 

EXAMPLES. 

1.  What  is  the  neat  weight  of  a  hhd.  of  Tobacco, 
weighing  15cwt.  3qrs.  201b.  gross,  tare  71b.  per  cwt.  tret 
and  cloff  as  usual  1  Cwt.  qrs.  ft.  oz. 

71b.  isTV)15     3     20     0  gross. 
3     27     8  tare. 


26)14     3     20     8  tare-suttle. 
2       85+  tret. 


168)14     1     12     3  tret-suttle. 
9     9+  cloff. 


14     1       2  10  Answer,  nearly. 

NOTE.  —  Some  say  21b.  for  every   lOOlb.   of  tret-suttle 
ought  to  be  allowed,  to  make  the  weight  hold  good  when 
•  sold  by  retail  ;  instead  of  21b.  on  every  3361b. 

2.  What  is  the  value,  at  5£d.  per  lb.,  neat  weight,  of 
26  chests  of  sugar,  each,  9cwt.  2qrs.  15Jflb.  gross,  tare 
131b.  per  cwt.,  tret  as  usual,  and  clofF  21b.  on  3001b.  ; 
251b.  a  qr.  ?  Ans.  <£499  15s.  5d. 


DOUBLE  RULE  OF  THREE. 

THE  DOUBLE  RULE  OF  THREE  teaches  to  solve  such 
questions  as  require  two  or  more  statings  in  the  Rule  of 
Three.  In  these  questions  there  is  always  given  an  odd 
number  of  terms  as  five,  seven,  or  nine,  &c.  These  are 
distinguished  into  terms  of  supposition,  and  terms  of  de- 
mand, the  number  of  the  former  always  exceeding  that 
of  the  latter  by  one,  which  is  of  the  same  kind  with  the 
term  or  answer  sought. 

RULE.  —  Write  the  term  of  supposition,  which  is  of  the 
same  kind  with  the  answer,  for  the  middle  term. 

Take  one  of  the  other  terms  of  supposition,  and  one  of 
the  demanding  terms  of  the  same  kind  with  it  ;  then 
place  one  of  them  for  a  first  term,  and  the  other  for  a 
third,  according  to  the  directions  given  in  the  Rule  of 
Three.  Do  the  same  with  another  term  of  supposition, 
and  its  corresponding  demanding  term  ;  and  so  on,  if 


112  DOUBLE    RULE    OF    THREE 

there  be  more  terms  of  each  kind,  writing  the  terms  un- 
der each  other,  which  fall  on  the  same  side  of  the  middle 
term. 

Multiply  together  all  the  terms  in  the  first  place,  and 
also  all  the  terms  in  the  third  place.  Then  multiply  the 
latter  product  by  the  middle  term,  and  divide  the  result 
by  the  former  product;  and  the  quotient  will  be  the  an- 
swer required.  Or,  take  the  two  upper  terms  and  the 
middle  term,  in  the  same  order  as  they  stand,  for  the  first 
stating  of  a  question  in  the  Single  Rule  of  Three  ;  then 
take  the  fourth  number  resulting  from  the  first  stating, 
for  the  middle  term  of  a  new  stating  in  the  above  Rule, 
and  the  two  under  terms  of  the  Double  Rule  of  Three 
stating,  in  the  same  order  as  they  stand,  for  the  extreme 
terms  of  the  new  stating  ;  and  the  fourth  term  resulting 
from  this  new  stating,  will  be  the  answer. 

NOTE  1.  —  The  first  and  third  terms  of  each  line,  if  of 
different  denominations,  must  be  reduced  to  the  same  de- 
nomination. 

2.  After  stating,  and  before  commencing  the  opera- 
tion, if  one  of  the  first  terms,  and  either  the  middle  term 
or  one  of  the  last  terms  will  exactly  divide  by  one  and 
the  same  number,  let  them  be  divided,  and  the  quotients 
used  instead  of  them  ;  which  will  much  shorten  the  work, 
Make  trial  with  the  first  Example. 
EXAMPLES. 

1.  How  many  men  can  complete  a  trench  of  135  yards 
long  in  eight  days,  provided  16  men  can  dig  54  yards  in 
6  days  ? 


54  yards)   :  ,6  men::  {  ^  Jg*  j   : 
8  days  )  f.  •  o  days  ) 

432  810 

16 

Here  54-5-27=2.  135-4-27=5 
8.5-2=4.     And  6-2=3  4860 

Then  2x4=8,  and  16-=-S=2.  810 

And  5x3=15,  arid  15x2=30. 

Therefore,  the  answer  by  432)12960(30  men,  Ans. 

Note  2,  is  30  men.  1296 

0 


DOUBLE    RULE    OF    THREE.  113 

2.  If  a£100  in  one  year  gain  ^6  interest,  what  will  be 
the  interest  of  ,£750  for  7  years  ? 

-•*.'*• 


. 

3.  A  farmer  sells  204  dollars'  worth  of  grain  in  5  years, 
when  it  sold  at  60  cents  per  bushel  ;  what  is  it  per  bushel 
when  he  sells  1000  dollars'  worth  in  18  years,  if  he  sell 
the  same  quantity  yearly?          Ans.  8lcts.  6T9^mills.+ 

4.  If  7  men  can  reap  84  acres  of  wheat  in  24  days,  how 
many  men  can  reap  1  00  acres  in  1  0  days  1  Ans.  20  men. 

5.  If  6  men  build  a  wall  20  feet  long,  6  feet  high  and 
4  feet  thick,  in  16  days  ;  in  what  time  will  24  men  build 
one  200  feet  long,  8  feet  high  and  6  feet  thick  T 

Ans.  80  days. 

24  men  }  f  6  men  } 

20  feet  long    f  ,  >  200  feet  long    f  .  go  , 

6  feet  high    (  '  '  )  8  feet  high        (  ' 

4  feet  thick  ;  (6  feet  thick       ) 

6.  An  usurer  put  out  75  dollars  at  interest,  and  at  the 
end  of  8  months  he  received  for  principal  and  interest* 
79  dollars  ;  I  demand  at  what  rate  per  cent,   he  received 
interest.  Ans.  8  per  cent. 

7.  If  the  carriage  of  13cwt.  Iqr.  for  72  miles  be  £Z 
10s.  6d.  what  will  be  the  carriage  of  7cwt.  3qrs.  for  112 
miles  ?  Ans.  ^2  5s.  lid.  1  -jZ&qr. 

S.  If  a  family  of  9  persons  spend  450  dollars  in  5 
months,  how  much  would  be  sufficient  to  maintain  them 
8  months,  if  5  more  were  added  to  the  family  ? 

Ans.  1120  dollars. 

9..  What  is  the  interest  of  654  dollars  for  164*days,  at 
6  per  cent,  per  annum  ?  Ans.  17  dolls.  63ets.  1m.+ 

10.  If  248  men  in  5  days  of  1  1  hours  each,  dig  a  trench 
230  yards  long,  3  yards  wide  and  2  deep  ;  in  how   many 
days  of  9  hours  long,  will  24  men  dig  a  trench  420  yards 
long,  5  wide  and  3  deep  ?  Ans.  28S^VV  days. 

11.  If  30  men  perform  a  piece  of  work  in  20  days,  how- 
many  men  will  effect  another  piece  of  work,  4   times  as 
large,  in  a  fifth  part  of  the  time  ?  Ans,  600  men. 

12.  Wlmt  principal  will  gain  §315  in  7  years,  at  6  per 
cent,  per  annum  ?  Ans.  $750. 

13.  If  SOOOJfe.  of  beef  will  serve  340  seamen  15  days, 
how  many  pounds  will  serve  120  seamen  25  days  T 

Ans.  1764ft. 


114  CONJOINED    PROPORTION. 

CONJOINED  PROPORTION. 

CONJOINED  PROPORTION  is  when  the  coins,  weights,  or 
measures  of  several  countries,  are  compared  in  the  same 
question  ;  or  it  is  the  joining  together  of  several  ratios, 
and  inferring  the  ratio  of  the  first  antecedent  and  last 
consequent,  from  the  ratios  of  the  several  antecedents  and 
their  respective  consequents. 

CASE  I. 

When  it  is  required  tojind  liow  many  of  the  last  kind  of 

coin,  weight,  or  measure,  mentioned  in   the  question,  are 

equal  to  a  given  number  of  thejirst. 

RULE  1.  —  Multiply  continually  together  the  antecedents 
for  the  first  term,  and  the  consequents  for  the  secondi  arid 
make  the  given  number  the  third. 

2.  Then  find  the  fourth  term  or  proportional,  which 
will  be  the  answer  required. 

EXAMPLES. 


1.  If  lOffe.  at  Boston  make  9ffo.  at  Amsterdam  ; 

at  Amsterdam  112j^.  at    Thoulouse,  how  many  Jfo.  at 
Thoulouse  are  equal  to  50}fc.  at  Boston  1 
Ant.  Con. 

10  :  9 

90  :         112 

900  :        1008  :  :  50  :  56ft.  Ans. 

2.  If  20  brasses  at  Leghorn  be  equal  to  10  varas  at  Lis- 
bon ;  61  varas  at   Lisbon  to  75  American  yards;  how 
many  American  yards  are  equal  to  100  brasses  at  Leg- 
horn ?  Ans.  61ff  yards. 

CASE  II. 

WJien  it  is  required  to  jftnd  how  many  of  the  Jirst  kind  of 
coin,  weight,  or  measure,  mentioned  in  the  question,  are 
equal  to  a  given  number  of  the  last. 
RULE.  —  Proceed  as  in  the  first  case,  only  make  the  pro- 

duct of  the  consequents  the  first  term,  and  that  of  the  an- 

tecedents the  second. 


BARTER.  115 

EXAMPLES. 

1.  If  20  ells  English  make   11   canes  at  Rome;  44 
canes  at  Rome,  136  brasses  at  Venice  ;  how  many  ells 
English  make  85  brasses  at  Venice  ? 

Ant.         Con. 
20     :       11 
44     :      136 

880         1496  :  880  : :  85  :  50  ells,  Ans. 

2.  If  41  U.  S.  bushels  make  26  hanegas  at  Cadiz  ;  39 
hanegas  at  Cadiz,  162  alquiers  at  Lisbon  ;  27  alquiers  at 
Lisbon,  5  sacks  at  Leghorn  ;  and  20  sacks  at  Leghorn, 
21  tuns  at  Copenhagen  ;  how  many  U.  S.  bushels  make 
36  tuns  at  Copenhagen  ?  Ans.  70fbus. 

3.  If  300  U.  S.  miles  make  77  miles  in   Germany  ; 
1771  miles  in  Germany,  1250  posts  in  France  ;  and  25 
posts  in  France,  38  miles  in  Holland  ;  how  many  U.  S. 
miles  make  80  in  Holland  ?       Ans.  290|£  U.  S.  miles. 


BARTER. 

BARTER  is  the  exchanging  of  one  commodity  for  anoth 
er,  and  directs  traders  so  to  proportion  their  goods,  that 
neither  party  may  sustain  loss. 

RULE.* — Find  the  value  of  that  commodity  the  quan- 
tity of  which  is  given  ;  then  find  what  quantity  of  the 
other,  at  the  rate  proposed,  you  may  have  for  the  same 
money,  and  it  gives  the  answer  required. 

EXAMPLES. 

1.  How  many  dozen  of  candles,  at  3s.  6d.  per  dozen, 
must  be  given  in  barter  for  4cwt.  2qrs.  of  tallow,  at  46s. 
per  cwt.  ? 

*  This  rule  is  only  an  application  of  the  Rule  of  Three. 


116  BARTER. 


qrs.       s.  cwt.  qrs. 

4     :  46     :  :     4     & 

18  4 

368          18 
46 

.    :  4)828 
2,0)20,7 


s.    d.  doz. 

36:        1     :: 

2.  A  buys  of  B  4  hogsheads  of  wine  containing  410 
gallons,  at  1  dollar  17  cents  per  gallon;  and  253  {fo.  of 
coffee  at  21  cents  per  ft-  :  In  part  of  which  he  pays  him 
21  dollars  in  cash,  and  the  balance  in  boards  at  8  dollars 
per  thousand  ;  how  many  feet  of  boards  does  the  balance 
require  ?  Ans.  63978J  feet. 

3.  Bought  a  sloop  of  70  tons  at   16  dollars  per  ton; 
paid  in  cash  500  dollars,  350  gallons  of  molasses  at  64cts. 
per  gallon,  and  the  balance  in  oil  at  74  cts.  per  gallon  ; 
how  many  gallons  did  it  amount  to? 

Ans.  535-gygaI. 

4.  A  barters  with  B  150  bushels  of  wheat  at  5s.  9d. 
per  bushel,  for  65  bushels  of  corn  at  2s,  lOd.  per  bushel, 
and  the  balance  in  oats  at  2s.  Id.  per  bushel  ;  what  quan- 
tity of  oats  must  A  receive  ?  Ans.  325f  bushels. 

5.  How  much  wine  at   1    dollar  28  cents  per  gallon, 
must  I  receive  in  barter  for  26cwt.  2qrs.  14j^.  of  raisins, 
at  9  dollars  44  cents  4  mills  per  cwt.,  25}^.  a  qr.  ? 

Ans.  196gals.  2qts.  l,704gills. 

6.  A  delivers  B  3  hogsheads  of  wine   at  6s.   Sd.  per 
gallon,  for  126  yards  of  cloth  ;  what  was  the  cloth  per 
yard,  in  Federal  Money  T  Ans.  1  dollar  66f  cts. 

7.  A  has  a  quantity  of  pepper*  weight  neat  1600}^.  at 
Is.  5d.  per  Jfo.  which  he  barters  with  B  for  two  sorts  of 
goods,  the  one  at  5d.  the  other  at  8d.  per  pound,  and  to 
have  ^  in    money,    and   of  each  sort  of  goods  an  equal 
quantity  ;  how  many  ffe.  of  each  must  he  receive,   and 
how  much  in  money  ? 

Ans.  I394fflb.  of  each,,  and  .£37  15s.  6£d, 


LOSS   AND    GAIN.  117 

8.  A  and   B  barter;    A  has    145   gallons   of  oil,   at 
$l,20cts.  per  gallon  ready  money,  but  in  barter  he  will 
have  $l,35cts.  per  gallon;  B  has  linen   at  58  cents  per 
yard  ready  money  ;  how  must  B  sell  his  linen  per  yd.,  in 
proportion  to  A's  barter  price,  and  how  many  yards  are 
equal  to  A's  oil  ? 

.        (  B's  linen  is  65cts.  2£m.  barter  price,  and 
5'  (      he  must  give  A  300yds.  for  his  oil. 

9.  K  and  L  barter  ;  K  has  woollen  cloth  worth  $1,33 
cts.  per  yard,  which  he  barters  at  $l,54cts.  with  L.,  for 
linen  cloth  at  50cts.  per  yard,  which  is  worth  43cts.  per 
yard  ;  —  who  has  the  advantage  in  barter,  and  how  much 
linen  does  L  give  K  for  70  yards  of  woollen  1 

.        (  215fyds.  of  linen,  and  L  has  the  advantage,  his 
5*  (      proportional  barter  price  being  only  49^J  cts. 

10.  J.  Tucker  and  Jonathan  Olmstead  barter  ;  the  for- 
mer gives  the  latter  90  gallons  of  wine  at  $1,28  per 
gallon  ;  for  which  the  latter  gives  the  former  10  guineas 
at  28s.  each,  in  money,  and  «500fJ).  of  cotton  ;  —  what  is  it 
valued  at  per  pound  ?  Ans.  13f|  cts. 

11.  Giles  Jackson  has  100  reams  of  paper,  at  $1,33£ 
cts.  ready  money,  which  in  barter  he  sets  down  at  $l,66f 
cts.    Robert  Howard,  sensible  of  this,  has  pamphlets  at  8£ 
cts.  apiece  ready  money,  which  he  adequately  charges,  and 
insists,  besides,  on  £  Of  the  price  of  those  he  parts  with, 
in  money  ;  —  what  number  of  the  books  is  he  to  deliver  in 
lieu  of  Jackson's  paper  ?    what  cash  will  make  good  the 
difference  1  and  how  much  is  Howard  the  gainer  by  this 
affair?     Ans.  1600  books  to  be  delivered;  $4l,6(3fcts. 
Howard  is  to  have  in  cash  ;  and  the  gain  to  Howard  is 
$41,<56fcts. 


LOSS  AND  GAIN. 

Loss  AND  GAIN  is  a  rule  that  discovers  what  is  gained 
or  lost  in  buying  or  selling  goods;  and  instructs  mer- 
chants and  traders  to  raise  or  lower  the  price  of  their 
goods,  so  as  to  gain  or  lose  a  certain  sum  per  cent. 

Questions  in  this  rule  are  performed  by  the  Rule  of 
Three. 


118       .  LOSS    AND    GAIN. 

There  is,  indeed,  great  variety  in  questions  in  this 
rule ;  but  they  may  be  all  easily  solved  by  a  little  consid- 
eration, and  the  following  proportion,  viz. — That  the 
gains  or  losses,  are  in  proportion  as  the  quantities  of  goods. 

EXAMPLES. 

1.  Bought  30  hogsheads  of  molasses  for  600  dollars  ; 
paid  in  duties  20  dollars  66  cents,  freight  40  dollars 
78  cents,  for  storage  6  dollars  5  cents,  and  for  insurance 
30  dollars  84  cents :  If  I  sell  it  at  26  dollars  per  hogs- 
head, how  much  shall  I  gain  per  cent.  ? 

$.  cts. 

600  $26 

20  66  30  hhds. 

40  78  

6  05  $780  00  Sold  for. 

30  84  698  33  Cost. 


698  33  81  67  Gain. 

$.  cts.         $.  cts.  $.  $.  cts. 

698  33    :    81  67       ::       100      :      11  69+Ans. 

2.  At  3s.  6d.  profit  on  the  pound,  how  much  per  cent.  ? 

Ans.  .£17  10s. 

3.  If  !{£.  of  coffee  cost  12cts.  and  it  sold  for  15  cents, 
what  is  the  profit  on  2931b.  neat  ?     Ans.  Sdolls.  79cts. 

4.  If  a  gallon  of  wine  cost  6s.  8d.  and  is  sold  for  7s. 
2d.  what  is  the  gain  per  cent.  ?  Ans.  7£  per  cent. 

5.  Sold  a  repeating  watch  for  175  dollars,  upon  which 
I  lost  17  per  cent,  whereas  I  ought  to  have  gained  20  per 
cent. ;  how  much  was  it  sold  for  under  its  just  value  ? 

Ans.  78dolls.  lct.+ 

6.  If  I  buy  broadcloth  for  13s.  5d.  per  yard,  how  must 
I  sell  it  to  gain  at  the  rate  of  25  per  cent.  ? 

£.  £.  s.    d.  s.    d. 

100      :      125      :  :      13  5     :      16  9f  Ans. 
Or  thus,  4)13  5 


16s.  9|d. 

7.  Bought  oil  for  90  cents  per  gallon  ;  at  what  rate 
must  it  be  sold  to  gain  20  per  cent.  1  Ans.  108cts. 


FELLOWSHIP.  119 

8.  Bought  115  gallons  of  wine  at  1  dollar  10  cents  per 
gallon  ;  how  many  gallons  of  water  must  be  put  in,  so  as 
to  gain  5  dollars  by  selling  it  at  1  dollar  per  gallon  1 

Ana.  16£  gallons. 

9.  Bought  a  hhd.  of  molasses  containing  119  gallons* 
at  52cts.  ger  gallon  ;  paid  for  carting  the  same  $l,25cts.  ; 
and  by  accident  9  gallons  leaked  out  ;  —  at  what  rate  per 
gallon  must  I  sell  the  remainder,  so  as  to  gain  $13  in  the 
whole  ?  Ans.  69cts.  2m.  + 

10.  Bought  llcwt.  of  sugar,  at  6£d.  per  ft-  but   could 
not  sell  it  again  for  any  more  than  £b  16s.  per  cwt.  ;  did 
I  gain  or  lose  by,  my  bargain  1     Ans.  Lost  <£2  11s.  4d. 

11.  Bought  cloth  at  17s.  6d.  per  yard,  which  on  exam- 
ination, I  find  to  be  much  damaged  ;  and  am,  therefore, 
content  to  lose  15  per  cent,  by  it  ;  —  how  must  I  sell  it 
per  yard  ?  Ans.  14s.  10|d. 

12.  By  selling  broadcloth  at  $3,25cts.  per  yard,  I  lose 
at  the  rate  of  20  per  cent.  ;  what  is  the  prime  cost  of  said 
cloth  per  yard  ?  Ans.  $4,6  cents,  2£m. 

13.  If,  when  I  sell  cloth  at  $7  per  yard,  I  gain  $10  per 
cent.  ;  what  will  be  the  gain  per  cent,  when  it  is  sold  for 
$8  per  yard  ?  Ans.  825,71  cents,  4m.-f 

14.  If  I  sell  a  cwt.  of  sugar  for  $8,  and  thereby  lose 
12  per  cent.  ;  what  shall  I  gain  or  lose  per  cent,  if  I  sell 
4cwt.  of  the  same  sugar  for  $36  1 

Ans.  I  lose  only  1  per  cent. 


FELLOWSHIP. 

FELLOWSHIP  is  a  rule  by  which  merchants,  &c.  trad- 
ing in  company  with  a  joint  stock,  determine  each  per- 
son's particular  share  of  the  gain  or  loss  in  proportion 
to  his  share  in  the  joint  stock. 

By  this  rule  a  bankrupt's  estate  may  be  divided  among 
his  creditors  ;  as  also  legacies  adjusted  when  there  is  a 
deficiency  of  assets  or  effects. 

SINGLE  FELLOWSHIP. 

Single  Fellowship  is  when  different  stocks  are  employ- 
ed for  any  certain  equal  time. 


120  SINGLE    FELLOWSHIP. 

RULE.* — As  the  whole  stock  is  to  the  whole  gain  or 
loss,  so  is  each  man's  particular  stock  to  his  particular 
share  of  the  gain  or  loss.f 

PROOF. — Add  all  the  shares  together,  and  the  sum  will 
be  equal  to  the  gain  or  loss,  when  the  work  is  right. 

EXAMPLES. 

1.  A  and  B  gained  by  trade  ,£182.     A  put  into  stock 
.£300  and  B  £400 ;  what  is  each  person's   share  of  the 
profit  ? 

£      £        £        £          £          £. 
300+400=700  :  182  :  :  300  :     78  A's  share. 
700  :  182  :  :  400  :  104  B's  share. 

£182  Proof. 

2.  A  man  dying,  bequeathed  his  estate  to  his  three  sons 
in  the  following  manner,  viz.  to  the  eldest  he  gave   1840 
dollars,  to  the  second  1550  dollars ;  and  to  the  third  960 
dollars  ;  but  it  was  found  his  whole  estate  was  no  more 
than  1840  dollars ;  what  is  each  one's  proportion  1 

(  $77S,29!f  the  first. 
Ans.  <     655,63^f  the  second. 
(    406,06jf  the  third. 

3.  A  and  B  companied  ;  A  put  in  450  dollars,  and  re- 
ceived f  of  the  gain  ;  what  did  B  put  in  1     Ans.  $300. 

4.  Three  merchants  freight  a  ship  with  wine  ;  A  load- 
ed 110  tuns,  B  97  tuns,  and  C  133  tuns.     In  a  storm  the 
seamen  were  obliged  to  throw  85  tuns  overboard  ;    how 
much  must  each  sustain  of  the  loss  1 

Ans.  A  27  £,  B  24£,  C  33£  tuns. 

5.  Three  men,  A,  B,  and  C,  contract  to  build  the  hull 
of  a  vessel  for  625  dollars  ;  A  works  100  days,   and   his 
work  is  estimated  at  1  dollar  80  cents  per  day  ;  B  works 
101  £  days,  estimated  at  1  dollar  60  cents  per  day  ;  and  C 

*  That  the  gain  or  loss  in  this  rule  is  evidently  in  proportion  to 
their  stocks,  may  be  shown  from  the  nature  of  the  Rule  of  Three. 

•f  The  first  and  third  contractions  of  note  4  of  the  Rule  of 
Three,  are  often  the  best  for  working-  questions  in  this  ruley 
especially  in  decimals. 


SINGLE    FELLOWSHIP.  121 

work  98  days,  at  1  dollar  50  cents  per  day  :  how  much 
is  each  man's  proportion  according  to  his  work? 

day.  $.  cts.     days.  day.  $.  cts.     days.  day.  $.  cts.  days. 
1  :  1    80  ::  100     1  :     I  60  :  :  101^  *    ^ 

100  1 


180,00  A's  work. 
162,00  B's  do. 
147,00  C's  do. 

147,00 
489  162,00 

$.          $.  $.  $•  cts. 

489    :    625    :  :    162     :     207,05^ 
489    :    625    :  :    147    :     187,88T% 
$.  $.  $.  $•     cts. 

489    :    625      :  :     180     :      230,06^ 

(  $230,06-^  A's  share. 
AnsJ    207,05TVvB's     do. 
(    187,8ST5e%  C's     do. 

$625,00  Proof. 

6.  A  ship  worth  3600  dollars  being  entirely  lost,  of 
which  ^  belonged  to  A,  £  to  B,  and  the  rest  to  C  ;  what  loss 
will  each  sustain  ?       Ans.  A  $450.  B  $900.  C  $2250. 

7.  A  and  B  gained  1260  dollars  of  which  A  is  to  have 
ten  per  cent,  more  than  B  :  what  is  the  share  of  each  ? 

Ans.  A  660dolls.     B  600dolls. 

8.  Three  merchants  made  a  joint  stock — A  put  in  ^565 
6s.  8d.  B  ,£478  5s.  4d.  and  C  a  certain  sum  ;  they  gained 
<£3739s.  lid.  of  which  C  took  ^112   Us.   lid.  for  his 
part ;  what  is  A  and  B's  part  of  the  gain,  and  how  much 
did  C  put  in  ?  (  A's  gain  £141  6s.  8d. 

AnsJ  B's  do.       119  Us.  4d. 
(  C  put  in     150  7s.  8d. 

9.  A,  B,  and  C,  traded  in  company ;  A  put  in  $140 ; 
B,  $250  ;  and  C  put  in    120yds.  of  cloth,  at  cash  price  ; 
they  gained  $230,  of  which  C  took  $100  for  his  share  of 
the  gain  ;— how  did  C  value  his  cloth  per  yard  in  com- 
mon stock,  and  what  was  A's  and  B's  part  of  the  gain  ? 

A       (  C  put  in  his  cloth  at  $2£  per  yard  ;  B's  part  of 
5t  \  the  gain  was  $83,33^cts. ;  and  A's  5546,66f  cts, 
mt 


, 


122  DOUBLE    FELLOWSHIP. 

DOUBLE  FELLOWSHIP. 

DOUBLE  FELLOWSHIP  is  when  the  stocks  are  employed 
for  different  times. 

RULE.*  —  Multiply  each  man's  stock  by  the  time  of  its 
continuance  ;  then  say,  as  the  sum  of  all  the  products  is 
to  the  whole  gain  or  loss,  so  is  each  man's  particular  pro- 
duct to  his  particular  share  of  the  gain  or  loss. 
EXAMPLES. 

1.  A  and  B  hold  a  piece  of  ground  in  common,   for 
which  they  pay  £36  —  A  put  in  23  oxen  for  54  days,  B  21 
oxen  for  70  days  ;  what  part  of  the  rent  must  each  man  pay  1 
-23x54  =  1242 
21x70=1470 


1470     :      19  10  3        B's. 


.£36  Proof. 

2.  Two  merchants  enter  into  partnership  for  16  months 
A  put  in  at  first  1200  dollars,  and  at  the  end  of  9  months 
200  dollars  more  ;  B  put  in  at  first  1500  dollars,  and  at 
the  expiration  of  6  months  took  out  500   dollars  —  with 
this  stock  they  gained  772  dollars  20  cents  ;  what  is  each 
man's  part  of  it  1 

Ans.  A's  $401  70cents—  B's  $370  50  cents. 

3.  A,  B  and  C,  enter  into  partnership  ;  A   put  in  85 
dollars  for  8  months,  B  put  in  60  dollars  for  10  months, 
and  C  120  dollars  for  3  months;  by  misfortune  they  lost 
41  dollars  ;  what  part  of  the  loss  must  each  man  sustain  ? 

Ans.  A's  part  $17.    B's  $15.    C's  £9. 

4.  W.  Thomas  and  N.  White  were  joint  tenants  of  a 
mill,  in  the  building  of  which  Thomas  laid  out  g!50,  and 
White  S270.     At  the  end  of  7  months,  Thomas  sold  his 
share  to  White  ;  and  at  the  end  of  the  first  year  White 
sold  the  mill.     They  then  made  a   settlement;  and,  the 
year's  profit  of  the  mill  being  ascertained  at  $260,  what 
was  each  man's  share  ? 

Ans.  Thomas's  $54,16|cts.     White's-  $205,83  Jets. 

*  When  the  times  are  equal,  the  shares  of  the  gain  or  loss  are 
evidently  as  the  stocks,  as  in  Single  Fellowship  ;  and  when  the 
stocks  are  equal  the  shares  are  as  the  times  ;,  but  when  neither 
are  equal,  the  shares  must  be  as  their  products. 


SIMPLE    INTEREST.  123 

5.  Jacob  M'Ewen,  Giles  Jackson,  John  Hastings,  and 
Anthony  Minot,  were  joint  tenants  of  a  certain  toll  bridge, 
which  they  held  for  the  terra  of  14  years,  by  charter. 
Their  whole  expense  in  building  the  bridge,  was  $'25745, 
50cts.,  of  which  M'Ewen  paid  $489(>,67cts.,  Jackson 
$1675,  Hastings  $12392,87cts.,  and  Minot  $6?8Q,96cts. 
At  the  end  of  2£  years,  M'Ewen  sold  out  to  Peter  Thom- 
son ;  at  the  end^of  5  years,  Jackson  sold  out  to  Jeremiah 
Apthorp  ;  and  at  the  end  of  10  years,  Hastings  sold  his 
share  to  James  Hawkins ;  at  the  expiration  of  the  14 
years,  the  whole  tollage  amounted  to  $30,000.  What 
was  each  man's  share  1 

'M'Ewen's,  $10l8,90c.  2m.+ 
Jackson's,       G97,07c.+ 
Hastings',    I03l4,87c.+ 


Answer. 


Minot's,  7901,52c.  8m.+ 
Thomson's,  4686,95c.  2m.+ 
Apthorp's,  1254,72c.  6m.+ 
Hawkins',  4125,94c.  8m.  + 


SIMPLE  INTEREST. 

INTEREST  is  the  sum  paid  by  the  borrower  to  the  lend- 
er for  the  use  of  money  lent. 

The  legal  interest  in  most  of  the  United  States  is  6 
per  cent,  per  annum  ;  that  is,  £6  for  the  use  of  <£lOO,  or 
$6  for  the  use  of  $100  for  one  year,  &c. 

Principal  is  the  money  for  which  the  compensation  is 
made. 

Rate  is  the  sum  per  cent,  agreed  on. 

Amount  is  the  principal  and  interest  added  together. 

Interest  is  of  two  kinds  ;  simple  and  compound.  Simple 
Interest  is  that  which  is  allowed  for  the  sum  lent  only. 

RULE.* — Multiply  the  principal  by  the  rate,  and  divide 
the  product  by  100;  and  the  quotient  is  the  interest  for 
one  year.  Multiply  the  interest  for  one  year  by  the  given 
number  of  years,  and  the  product  is  the  interest  for  that 
time.  For  any  parts  of  a  year,  as  months,  days,  &c.  di- 

*  Simple  Interest  is  only  anr  application  of  the  Rule  of  Three. 


124 


SIMPLE    INTEREST. 


vide  the  interest  for  one  year  by  the  aliquot  parts  of  a 
year  or  month ;  or  the  interest  may  be  found  by  a  state- 
ment in  the  Rule  of  Three.* 

NOTE. — In  Federal  Money  the  quotient  after  the  divis- 
ion by  100  gives  the  answer  in  the  same  name  with  the 
lowest  denomination  in  the  principal. 


TABLE  OF  ALIQUOT  PARTS. 


Parts  of  a  Year. 
6  Months  is  -  - 
4  -  -  - 

3        -  - 

2    -        - 


Parts  of  a  Month. 

3  Days  is 

o  - 

6         -       - 
10     -      - 
15      *         - 


TV 
* 
* 
* 


EXAMPLES. 

1.  What  is  the  interest  of  J6639  for  one  year,  at  6  per 
cent.?  Ans.  ^38  6s.  9£d.+ 

639 
6 


qr.  2,40 


*  Calling  30  days  a  month  gives  the  interest  too  much.  If  the 
principal  be  small,  the  errour  is  trifling-.  But,  if  the  sum  be 
very  large,  say— as,  365  days  :  is  to  the  interest  for  1  year  :  so 
is  the  given  number  of  days  ;  to  me  interest  required. 


SIMPLE    INTEREST.  125 

2.  What  is  the  interest  of  372  dollars  for  one  year,  5 
months,  and  5  days,  at  6^  per  cent.  ? 
372 


2232 
186 

4  months,  £      $24,18  Interest  for  one  year. 

1  month,  TL          8,06  Interest  for  four  months. 

5  days,  |  of  1  mo.  2,01^  Interest  for  one  month. 

do.     five  days. 


834,59+  Answer. 

3.  What  is  the  amount  of  <£49  6s.  4^d.  for  one  year, 
at  6  per  cent.  ? 

£.     s.     d. 
49     6     4 


2,95  18     3 

20 

s.  d. 

19,18  £49    6  4£  Principal. 

12  2  19  2    Interest. 


2,19       Ans.  =£52  5  6£  Amount. 

4.  What  is  the  interest  of  1600  dollars  for  one  year 
and  three  months,  at  6  per  cent.  ?  Ans.  $120. 

5.  What  is  the  interest  of  <£71  7s.  6£d.  for  2  years,  at 
6  per  cent.  ?  Ans.  £8  11s.  3^d.+ 

6.  What  is  the  interest  of  67  dollars,   62  cents  for  3 
years  and  2  months,  at  6  per  cent.  ? 

Ans.  $!2,84cts.  7miHs.+ 

7.  How  much  is  the  interest  of  325  dollars  for  3  years, 
at  6  per  cent.  ?  Ans.  $58,  50cts. 

8.  How  much  is  the  Jnterest  of  66  cents,  4  mills  for 
one  year  and  7  months,  at  6  per  cent.  ? 

Ans.  6cts.  3mills.+ 

9.  What  is  the  interest  of  48  dollars,  25  cents,  5  mills 
for  5  years,  at  5  per  cent.  ?       Aris*  $12,6cts.  3mills.+ 

L  2 


SIMPLE    INTEREST. 

10.  What  is  the  interest  of  48  dollars  for  3  years,  at  9 
percent.!  Ans.  $12,96cts. 

11.  What  is  the  interest  of  «£5  16s.  3d.  for  one  year 
and  6  months,  at  6  per  cent.  ?  Ans.  10s.  5jd.+ 

12.  What  is  the  interest  of  9672  dollars,  from  Nov. 
30,  1823,  to  July  4,  1826,  at  8  per  cent.  ? 

Ans.  $2007,42cts.  7ns. + 

To  find  the  interest  of  any  sum  at  6  per  cent,  per  annum, 
for  any  number  of  months. 

RULE.* — Multiply  the  principal  by  half  the  number  of 
months,  and  that  product  divided  by  100  will  be  the  inter- 
est for  the  given  time. 

EXAMPLES. 

1.  What  is  the  interest  of  64  dollars,  50  cents  for  8 
months^  at  6  per  cent.  ? 

64,50 

8-^-2=4  4 

1,00)2,58,00  Ans.  2  dolls.  58  cents, 

2.  How  much  is  the  interest  of  36  dollars,  84  cents  for 
5  months,  at  6  per  cent.  ?  Ans.  92cts.  1  mill. 

3.  How  much  is   the   interest   of  750   dollars  for    15 
months,  at  6  per  cent.  1  Ans.  $56,25cts. 

4.  What  is  the  interest  of  «£24  15s.  4£d.  for  10  months, 
at  6  per  cent.  I  Ans.  £~l  4s,  9d.  y^qr. 

5.  What  is  the  interest  of  468  dollars  for  one  month, 
at  6  per  cent,  1  Ans.  $2,34cts. 

To  find  the  intertst  of  any  sum  for  any  number  of  days 

when  the  rate  is  6  per  cent. 

RULE. — Multiply  the  principal  by  the  number  of  daysr 
and  divide  the  product  by  6083  ;  the  quotient  will  be  the 
interest  required. 

*  THE  REASON  OF  THE  RULE. — When  the  time  is  months,  mul- 
tiplying- by  the  rate  for  the  time  gives  the  answer. 

This  rate  is  found  by  multiplying-  the  time  by  the  given  rate 
per  cent,  for  a  year,  and  dividing-  the  product  by  12  ;  the  quotien-t 
is  the  rate  required,  and  is  always  equal  to  half  the  months,  when 
the  yearly  rate  is  6  per  cent 


SIMPLE    INTEREST.  127 

EXAMPLES. 

1.  What  is  the  interest  of  376  dollars,  20  cents  for  80 
days,  at  6  per  cent.  1         376,20 

80 

$  cts.  m. 

6083)30096,00(4,94  7+  Answer. 

2.  What  is  the  interest  of  £749  10s.  6d.  for  12  days,  at 
6  per  cent.  1  £.     s.     d. 

749  10    6 
12 


6083)8994  6  ()(<£!  9s.  6fd.+  Answer. 

SHORT  PRACTICAL  RULES 

For  calculating  Interest  at  6  per  cent,  either  for  months, 
or  months  and  days. 

I.    FOR    POUNDS,    SHILLINGS,    &C. 

RULE. — If  the  principal  consists  of  pounds  only,  cut  off 
the  unit  figure,  and  as  it  then  stands,  it  will  be  the  inter- 
est, for  one  month,  in  shillings  and  decimal  parts.  ^Mul- 
tiply this  sum  by  the  given  number  of  months,  and  take 
parts  for  the  days,  and  you  will  have  the  answer  in  shil- 
lings and  decimal  parts,  which  you  may  reduce.  But  if 
the  principal  consist  of  pounds,  shillings,  &c.  reduce  it 
to  its  decimal  value  ;  then  remove  the  decimal  point  one 
place  or  figure,  further  towards  the  left  hand,  and  as  the 
number  then  stands,  it  will  show  the  interest  for  one 
month,  in  shillings  and  decimals  of  a  shilling.  Multiply 
by  the  given  time,  &c.  and  you  will  have  the  answer. 

EXAMPLES. 

1.  Required  the  interest  of  ,£48  for  5  months  and  10 
days,  at  6  per  cent. 
Days.  s. 

!0|£|4,  8  interest  for  1  month* 
o 

24,  0     do.     for  5  nionths. 
1,6     do.     for  10  days, 

Ans.  25,  6  shillings  =£1  5s. 
12 

7,2 


128  SIMPLE    INTEREST. 

2.  Required  the  interest  of  £56  10s.  for  11  months,  at 
6  per  cent. 

£.  s.      £. 

56  JO  — 56,5  decimal  value. 
Therefore  5,65  shillings  interest  for  one  month, 
11 


Ans.  62,15  interest  for  11  months==£3  2s.  l,8d. 

3.  What  is  the  interest  of  .£37  8s.  6d.  for  one  year,  7 
months,  and  a  half,  at  6  per  cent,  per  annum  ? 

Ans.  £8  10s.  4d.  2,3qrs. 

4.  Required  the  interest  of  £28  16s.  for  20  days,  or 
two-thirds  of  a  month,  at  6  per  cent.    Ans.  Is.  ll,04d. 

NOTE. — The  foregoing  rule  will  serve  also,  when  the 
rate  is  5  per  cent,  or  7  per  cent.  For,  when,  by  the  Rule, 
you  have  found  the  interest  of  any  sum,  at  6  per  cent,  if 
you  decrease  the  said  interest  by  one-sixth  of  itself,  it 
then  becomes  the  interest  of  the  same  sum,  at  5  per  cent. ; 
or  if,  after  finding  the  interest,  by  the  Rule,  at  6  per 
cent.,  you  increase  the  said  interest  by  one-sixth  of  itself, 
it  then  becomes  the  interest  of  the  same  sum,  at  7  per  cent. 

Take  for  examples  the  first  two  of  the  preceding  ones. 

Required  the  interest  of  ^48  for  5  months  and  10  days, 
at  5  per  cent.  s. 

•J)25,6  interest  at  6  per  cent. 
4,26 


Ans.  21,33s.  int.  at  5  per  cent=^l   Is.  4d. 
What  is  the  interest  of  <£56  10s.  for  11   months,  at  7 
per  cent.  ? 
s. 

£)62,15  int.  at  6  per  cent. 
10,3583 

Ans.    72,5083  sbill.  int.  at  7  per  cent.=£3  12s.  6, Id. 

II.  For  Federal  Money. 

RULE. — Divide  the  Principal  by  2,  placing  the  separa- 
trix  as  usual,  and  the  quotient  will  be  the  interest  for  one 
month,  in  cents,  and  decimals  of  a  cent ;  that  is»  the 


SIMPLE    INTEREST.  129 

figures  at  the  left  of  the  separatrix  will  be  cents  and 
those  on  the  right,  decimals  of  a  cent.  Then  multiply 
this  sum  by  the  given  number  of  months,  or  months  and 
decimal  parts  thereof,  or  for  the  days  take  even  parts  of 
30,  &c. 

EXAMPLES. 

1.  Required  the  interest  of  $582,64cts.  for  5^  months, 
at  6  per  cent. 

2)562,64 

281, 32  int.  for  1  month. 


140,66  int.  for  £  month. 
1406,60  int.  for  5  months. 

Ans.     1547,26= l547T2^cts.«$l5347cts.  2,6m. 
Or  thus  :       281,32  int.  for  1  month* 
X5,5  months. 

1406  60 
14066  0 


1547,260cts.=$l5,47cts.  2,6m. 

2.  What  is  the  interest  of  $l8,48cts.'  for  3  years,  7 
months,  and  10  days,  at  6  per  cent.  ? 

2)18,48 

10  days=i  9,24  int.  for  1  month. 
43  months. 

2772 
3696 

897,32  int.  for  43  months. 
3,08  do.  for  10  days. 

Ans.     400,4,0cts.=$4  and  4  mills. 

3.  What  is  the  interest  of  $468  for  7  months,  at  6  per 
cent.?  Ans.  $l6,38cts. 

For  5  or  7  per  cent,  proceed  according  to  the  direc- 
tions in  the  preceding  Note. 


130  INTEREST    ON    NOTES,    &,C. 

To  compute  Interest  on  Notes,  Bonds,  fyc.,  having  par- 
tial Payments  or  Endorsements  thereon. 

RULE  1. — Cast  the  Interest  upon  the  whole  Principal, 
for  the  whole  Time  ;  then  separately  upon  each  Endorse- 
ment, for  its  respective  Time  ;  and  subtract  the  whole 
Amount  of  the  one  from  that  of  the  other. 

NOTE. — I  call  30  days  a  month,  and  12  months  a  year. 

EXAMPLES. 

1.  A  Note  dated  January  1st,  1821,  was  given  for 
$120,  interest  at  6  per  cent,  and  there  were  the  follow- 
ing payments  endorsed  upon  it : 

June  1,  1821,  rec'd  on  the  within  note  $50. 
Oct.  1,  1821,  rec'd  $40. 

I  demand  how  much  was  due  on  said  note,  Jan'y  1,  1822, 
when  it  was  taken  up  ? 

$ 

120  Principal,  or  sum  of  the  note. 
7,20cts.  Interest  for  the  whole  time. 


$127,20cts.  Amount. 

9.  8. 

50  First  payment.  40  Second  payment. 

l,75cts.  Interest.  0,60cts.  Interest. 


51,75  Its  Amount.         40,60  Its  Amount. 


9 .  c.  9.    c. 

51,75  )  Several  Am'ts       127,20  Amount  of  Note. 

40,60  }  of  Payments.          92,35  Amount  of  Pay'ts. 


92,35  Total  Amount.         $34,85cts.  Due  on  the  Note, 

Jan.  1,  1822,  Ans. 

2.  A  Bond  dated  April  17th,  1817,  was  given  for  $675, 
interest  at  6  per  cent.,  and  it  was  endorsed  as  follows  : 

May  7,  18:8,  rec'd  on  the  within  $148. 

August  17,  1820,  rec'd  $341. 

Jan.  2,  1822,  rec'd  $99. 

I  demand  what -was  due  on   said  bond,  June  17,   1822, 
when  it  was  paid  up  ?  Ans.  $2l9,52cts. 


INTEREST    ON    NOTES,    &C.  131 

3.  A  bond  or  note  was  given,  June  4th,  1819,  for  g699, 
interest  at  6  per  cent.  ;  it  was  endorsed  $78,  July  9,  1820: 

$147,  Nov,  27,  1821;  $6S,  Jan.  17,  1822;  and  $400, 
Jan,  30,  18*23  ; — I  demand  how  much  will  be  due  on  said 
note  May  30,  1823?  Ans.  $l32,87cts.  1m. 


£62  10s. 

4.  Value  received,  I  promise  to  pay  George  Appleton, 
the  sum  of  Sixty-two  Pounds,  Ten  Shillings,  and  interest 
at  6  per  cent,  per  annum,  till  paicl. 

PETER  FRISBIE. 

Halifax,  April  4,  1820. 

Endorsed,  £50,  Sept.  4,  1822. 

If  he  endorse  j(2l2  10s.  June  1,  1823,  how  much  will 
be  due  on  said  note,  December  4,  1823? 

Ans.  £9  12s. 


Contraction  of  the  Rule  of  6  per  cent. 

RULE.  —  Point  off  the  right  hand  figure  of  each  princi- 
pal for  a  decimal  ;  multiply  each  by  its  particular  time, 
and  add  the  products.  If  the  principals  be  pounds,  &c. 
the  sum  total  will  be  shillings  and  decimals  ;  but  if  Fed- 
eral Money,  halve  it  for  dimes,  interest:  which  added  to 
the  last  principal,  gives  the  sum  due  on  the  note. 

Take  tliejirst  Example  in  the  preceding  Rule. 

1st  Principal,         $12,&x5  months.  =60,0 
2d         do.  7,0x4  =28,0 

3d         do.  8,0x3  =09,0 


As  it  is  Federal  Money,  divide   by   2)97,0 

*  48,5dim.=48& 

dimes=$4,85cts.  which  added  to  the  last  principal, 
30+4,85  ~£34,S5cts.  amount  due.  Aris. 

This  is  merely  the  same  Rule  as  that  at  pages  128 — 129. 


132  INTEREST    ON    NOTES,    &C. 

For  an  Example  in  pounds,  fye.  take  the  last  question  in 
the  preceding  Rule. 

1st  Principal  £6%  10s.=6,25s.  by  Rule,  page  127. 
Yrs.     m.     d. 
1822    9      4 
1820    4      4 


250  12  10  =2d  Principals  1,25s. 

12 


Yrs.     m.     d. 


29  months.  1823    6 
s.        m.     s.        1822     9      4 

Int.  for  Imo.=6,25x29  =  181,25 m. 

Int.  for  Imo.=l,25x8,9=  11,125        8    27=8,9 


192,375  shillings, 
which  =£9  12s.  4£d.  Ans. 

There  being  no  principal  left  to  which  to  add  the  above 
sum,  that  sum  is  itself  the  exact  answer. 

This  is  the  same  Rule  as  that  at  page  127. 

Note. — If  the  last  principal  be  overpaid  by  the  last 
endorsement,  take  the  sum  thus  overpaid,  multiply  it,  in 
the  same  way,  by  the  time,  from  that  endorsement,  up  to 
the  settlement,  then  add  its  dimes,  interest,  to  itself,  and 
subtract  the  sum  from  the  interest  of  the  Note's  princi- 
pals, as  above  found. 

3.  Gave  a  note  for  $500,  Jan.  16,  1822,  payable  Nov. 
16,  1826 ;  $50  were  endorsed  thereon  April  1,  1823; 
$400,  July  16,  1824;  $(iO,  Sept.  1,  1825;  how  will  the 
balance  be  at  the  date  when  payable  1 

$500  =  1  st  principal.  500— 50=450=2d  principal. 
450 — 400=50=3d  principal.  Then  the  next  endorse- 
ment overpays  the  sum  borrowed.  60=3d  endorsement, 
and  60 — 50,  the  3d  principal =$10  overpaid  on  the  bum 
borrowed. 


INTEREST    ON    NOTES,    &C.  133 

mo.  mo. 

1st  prin.  $50,0x14,5=725,0  Overp'd  $1,0x26,5=26,5 
2d     do.      45,0x15,5=697,5     And  2)26,5 
3d     do.       5,0x13,5=  67,5 

-  Dimes 
2)1490,0         Add  sum  10,00 
Then,  $74,50 


1  J  /J&j-  subtr.    745  dimes,  $  1  1  ,32£ 

or,   §74,50ct.s.        -- 
Aus.  863,  17£  due  on  Note. 

4.  A  note  was  given,  May  10,  1821,  for  $685,  interest 
at  6  percent.  Nov.  15,  1821,  it  was  endorsed  $150  ;  April 
30,  1822,  g275  ;  Dec.  1,  1822,  $90  ;  July  30,  1823,  S45; 
March  10,  1824,  gf50,  and  Aug.  20,  1825,  it  was  paid  up  ; 
what  was  the  last  payment]  Aris.  $l£6,96^cts. 

RULE  2.  —  The  Supreme  Court  of  Massachusetts,  estab- 
lished the  following  more  equitable  Rule.  —  When  there 
are  endorsements,  find  the  interest  on  the  principal  to 
the  time  of  the  first  payment,  add  it  to  the  principal, 
and  from  the  sum  subtract  the  payment  then  made  ;  if 
the  endorsement  be  not  equal  to  the  interest  then  due, 
cast  the  interest  to  the  next  endorsement,  add  the  two 
endorsements  together,  and  from  the  amount  of  the  prin- 
cipal to  that  time,  subtract  their  sum  ;  on  the  remainder 
cast  the  interest  to  the  time  of  the  next  endorsement, 
subtracting  each  payment  as  you  proceed,  if  it  be  not 
less  than  the  interest  then  arisen  on  the  principal  sum  ; 
and  so  on  to  the  date  of  giving  up  the  obligation. 

EXAMPLES. 

I.  A  note  was  given,  January  20th,  1821,  for  $360, 
SOcts;  September  10th,  following,  200  dollars  were  en- 
dorsed, and  December  20th,  1822,  it  was  taken  up  ;  what 
was  the  last  payment,  interest  at  6  per  cent.  ;  computing 
it  by  both  rules,  and  showing  their  difference  ? 
Yr.  m.  d.  Yr.  m.  d. 

1821      9      10=Sep.  10.  1822  12    20=Dec  20. 

1821      1      20=Jan.  20  1821     9    10=Sept  10. 

7      20  1     3    10 

M 


134  INTEREST    ON    NOTES,    &C. 

$.    Cts. 

360,50      Note  dated  Jan.  20,  1821.       mo.  d. 
13,81  9  Interest  up  to  Sept.  10,  1821=7  20 

374,  31  9  Amount. 

200, 00  0  First  payment  deducted. 


174,  3i  9  Due  Sept.  10,  1821.  Yr.  m. 

13,  36  4  Interest  to  Dec.  20,  1822=1  3£ 


$187,  68  3  Last  payment  by  Rule  2. 

Yr.  m.  d. 

1822  12  20 

1821     1  20 


1   11     0 

$.  cts. 

360,  50      Note.  Yr.  m. 

41,  45  7  Interest  up  to  Dec.  20,    1822=1     11 


401,  95  7  Amount  for  the  whole  time. 


200,00  0  First  payment,  Sept.  10,  1821.  Y.  m.  d. 
15,  33  3  Interest  up  to  Dec.  20,     1822=1  3    10 


215,33  3  Amount. 

$.     cts.  m. 

401,    95  7  Amount  of  note. 
215,    33  3  Amount  of  payment. 


186,  62   4  Last  payment  by  Rule  1. 

$.     cts.  m. 

187,  68    3  Due  by  Massachusetts'  Rule. 
186,  62    4  Due  by  common  Rule. 


1,  05   9  Difference. 

2.  A  note  was  given,  Nov.  15,  1820,  for  g282,  56cts. ; 
May  10,  1821,  g96,34cts.  were  endorsed  ;  December  20, 
following,  §174,28ets. ;  May  10,  1822,  $lO,50cts.;  and 
November  15,  following,  $ 5,25cts. ; — what  will  be  due  on 


COMPOUND    INTEREST.  135 

said  note  June  10,  1823,  reckoning  interest  at  6  per  cent., 
computing  by  both  rules,  and  showing  their  difference  1 

f  By  Massachusetts'  rule     $13,  164 
Answer.  J  By  ru,e  firgt 


Difference  $1,  709 

3.  An  obligation  was  given,  July  15,  1817,  for  $340,  in- 
terest at  6  per  cent.,  on  which  are  the  following  payments  ; 
Dec.  25,  following,  $7,50cts.  ;  June  10,  1818,  $iO,25cts.; 
January   1,   1819,  $15;    August    16,    1820,    825,75cts.  ; 
November  1,  1821,  $30  ;  and  December  25,  1822,  §250  ; 
if  it  be  taken  up  March  25,  1823,  what  sum   will  cancel 
it,  by  the  Massachusetts'  rule  ?  Ans.  $113,  53cts. 

N.  B.  —  The  figures  beyond  mills,  in  the  two  preceding 
examples,  are  all  omitted  in  casting. 

4.  If  a  note,  given  July  5,  1  819,  for  $101,  at  6  per  cent., 
were  endorsed  July  5,  1820,  86  ;  July  5,  1821,  §6;  July 
5,  1822,  §6;  July  5,  1823,  $6;  July  5,  1824,   $6;  July 
5,  1825,  86;  July  5,  1826,    $6;  July  5,  1827,    $6;  July 
5,  1828,  $6;  what  would  be  due,  July  5,   1829,  by  both 
rules  and  how  much  more  by  the  second  rule  than  by  the 
first  ; 

r$  107,60  by  Mass,  rule;    $91,40   by  common 
Ans.  <  rule  ;  and  816,20  more  by  the  former  than   the 
(  latter. 


COMPOUND  INTEREST. 

COMPOUND  INTEREST  is  what  arises  from  the  interest 
being  added  to  the  principal,  and  becoming  a  part  of  the 
principal,  at  the  end  of  each  stated  time  of  payment. 

RULE.  —  Find  the  Simple  Interest  of  the  given  sum  for 
one  year,  or  the  time  of  the  first  payment  ;  add  it  to  the 
principal,  and  find  the  interest  of  the  amount  for  the  next 
year  or  payment,  and  so  on  for  the  number  of  payments 
required.  Subtract  the  principal  from  the  last  amount, 
and  the  remainder  will  be  the  compound  interest. 


136  COMPOUND    INTEREST. 

EXAMPLES. 

1.  What  is  the  compound  interest  of  406  dollars,  for  3 
years,  at  6  per  cent,  per  annum  ? 

406  principal  for  the  1st  year. 
6 


24,36  interpst  of  do.  > 

40(5,      principal  for  the  1st  year.   } 

430,36  principal  for  the  2d  year. 
6 

25.82,1,6  interest  of  do.  )      , 

430,36        principal  for  the  2d  year.    /  * 

456,18,1     principal  for  the  3d  year. 
6 


27,37,0,86       interest  for  do.  >  A 

456,18,1        principal  for  the  3d  year.         J  A 

483,55,1      amount  for  three  years. 

406,  principal  for  the  1st  year,  subtracted. 

Ans.  $77,55,1    compound  interest. 

2.  How  much  is  the  compound  interest  of  2535  dollars 
for  four  years,  at  6  per  cent,  per  annum  ? 

Ans.  665  dolls.  36cts.+ 

N.  B. — The  mills  are  here  all  rejected  in  casting. 

3.  What  is  the  compound  interest  of  1000  dollars  for  5 
years,  at  6  per  cent.  ?       Ans.  338dolls.  22cts.  4mills.+ 

N.  B. — The  figures  beyond  mills  are  here   omitted  in 
casting.     The  next  is  done  in  whole  numbers. 

4.  What  is  the  compound  interest  of  ^£128  17s.  6d.  for 
6  years,  at  6  per  cent.  1  Ans.  .£53  18s.  8d.+ 

5.  How  much  will  680  dollars  amount  to  in  4  years,  at 
6  per  cent,  compound  interest  ? 

Ans.  S858,48cts.  3m.  + 

N.  B. — Figures  beyond  mills  omitted. 


COMMISSION. INSURANCE.  137 


COMMISSION.* 

COMMISSION    AND    BROKERAGE  are  compensations  to 
factors  and  brokers  for  their  respective  services. 

EXAMPLES. 

1.  What  is  the  commission  on  4760dolls.  at  2J-  pe-r 
cent.  ? 

2)4760 


9520 
2380 

119,00  Ans.  119  dollars. 

2.  What  is  the  commission  on  ,£526  11s.  5d.  at  3J-  per 
cent.?  Ans.  £IS  8s.  7d.+ 

3.  What  is  the  brokerage  on  926  dollars,  59  cents,  at 
1£  per  cent.  1  Ans.  13dolls.  89cts.  7m.-J- 

4.  What  is  the  commission  on  1293  dollars,  53   cents, 
at  I  per  cent.  1  Ans.  9dolls.  73cts.  8mills,+ 

5.  Required    the    neat    proceeds    of    certain    goods 
amounting  to  2176  dollars,  deducting  a  commission  of  -£ 
per  cent.  ?  Ans.  2i5(idolls.  96cts. 

6.  A  factor  receives  3690  dollars  to  lay  out  in   potash, 
reserving  from  it  his  commission  of  2^   per  cent,   on  the 
purchase;  the   potash   being  190  dollars   per  ton,   how 
much  did  he  purchase  ?  Ans.  18tons,  IScwt.  3qrs.  22T2Fft. 


INSURANCE. 

INSURANCE  is  an  exemption  from  hazard,  by  paying  a 
certain  sum  on  condition  of  being  indemnified  for  loss  or 
damage  of  ships,  houses,  merchandise,  &c.  which  may 
happen  from  storms,  fires,  &c. 

*  The  method  of  working1  questions  in  this  and   the  following 
rules  of  Insurance,  &c.  is  the  same  as  in  Simple  Interest. 
M2 


138  DISCOUNT. 

EXAMPLES. 

1.  What  is  the  premium  of  insuring  8250  dollars,  at  6 
per  cent.  1 

8250 
6 


495,00  Ans.  495  dollars. 

What  is  the  premium  of  insuring   1650  dollars,  at 
per  cent.  ?  Ans.  255dolls.  75cts. 

3.  What  sum  must  be  received  for  a  policy  of   1658 
dollars,  deducting  a  premium  of  23  per  cent,   for  insur- 
ance ?  Ans.  1276dolls.  66cts. 

4.  What  is  the  premium  for  the  insurance  of  4000  dol- 
lars, at  7f  per  cent.  ?  Ans.  305  dollars. 

5.  What  sum  must  be  insured  upon  to  cover   1800 
dollars,  when  the  premium  is  10  per  cent.  ? 

100  Policy. 
Deduct    10  Premium. 

90  Sum  covered. 
If  $90  :  $100     :  :     $1800  :  $2000  Ans. 


DISCOUNT. 

DISCOUNT  is  an  allowance  made  for  the  payment  of  any 
sum  of  money  before  it  becomes  due  ;  arid  is  the  differ- 
ence between  that  sum  due  some  time  hence,  and  its 
present  worth.  The  present  worth  of  any  sum,  due  some 
time  hence,  is  such  a  sum,  as,  if  put  to  interest,  would  in 
that  time,  and  at  the  rate  per  cent,  for  which  the  discount 
is  to  be  made,  amount  to  the  sum  or  debt  then  due. 

RULE.  —  As  the  amount  of  100  dollars  for  the  given  rate 
and  time,  is  to  the  interest  of  100  dollars  for  that  time,  so 
is  the  given  sum  or  debt,  to  the  discount  required.  Or, 
as  the  amount  of  100  dollars  or  pounds,  is  to  100,  so  is 
the  given  sum  or  debt,  to  the  present  worth  required. 

NOTE.  —  When  goods  are  bought  or  sold,  money  ad- 
vanced, bank-bills  exchanged,  &c.  and  discount  is  to  be 


DISCOUNT.  139 

made  at  any  rate  per  cent,  without  time,  the  interest  of 
the  sums  as  found  for  a  year,  is  the  discount. 

EXAMPLES. 

1.  What  is  the  discount  of  1912  dollars,  50  cents  due 
3  years  hence,  at  4£  per  cent.  1 

4,50 
3 

13,50 
100, 

g.  cts.  $.  cts.       8.  cts. 

$113,50  :  13,50     :  :      1912,50  :  227,47+Ans. 

2.  What  is  the  present  worth  of  760  dollars  due  in  8 
months,  discount  at  6  per  cent,  per  annum  ? 

8. 

8  mo.  46=4 
100 

8.  8. 

104     :       100     ::     760   : 

Ans.  730dolls.  76cts.  9mills.+ 

3.  what  is  the  present  worth  of  500  dollars  payable  in 
^  of  a  year,  discount  being  at  5  per  cent.  1 

Ans.  $493,S2fcts.+ 

NOTE. — When  several  sums  are  to  be  paid  at  various 
times,  find  the  discount  or  present  worth  of  each  sum  sep- 
arately, and  then  add  those  discounts  of  present  worths 
into  one  sum,  in  order  to  obtain  the  required  answer. 

4.  A  is  to  pay  592  dollars,  70  cents  on  the  first  day  of 
April,  1833,  and  598  dollars,  90  cents   the  first  of  July 
following.     It  is  required  to  know  how  much  money  will 
discharge  both  sums  on  the  first  of  January,   1833,   dis- 
counting at  8  per  cent,  per  annum. 

Ans.  H56dolls.  94cts.  3mills.+ 

5.  Bought  a  quantity  of  goods  for   500  dollars  ready 
money,  and  sold  them  again  for  666  dollars,  67  cents,  pay- 
able at  f  of  a  year ;  what  was  the  gain  in  ready  money, 
supposing  discount  to  be  made  at  5  per  cent.  ? 

Ans.  142dolls.  57cts.+ 

6.  How  much  ready  money  will  discharge  a  note  for 
150  dollars  due  in  60  days,  allowing  6  per  cent,  per  an- 
num discount!  Ans.  148dolls.  51cts.  4mills.+ 


140  ANNUITIES. 

7.  If  a  legacy  of  2000  dollars  be  left  to  me  ;  500  dol- 
lars payable  in  6  months  ;  800  in  one  year ;  and  the  rest 
at  the  end  of  3  years  ;    and  the  executor  be   willing  to 
make  me  present  payment,  discounting  at  6  per  cent. ; 
what  ought  I  to  receive  1    Ans.  183;3dolls.  37cts.  4m.  + 

8.  What  is  the  present  worth  of   £60,  payable   at  3 
and  6  months,  at  5  per  cent,  per  annum  discount  ? 

Ans.  c£58  17s.  lid. 


ANNUITIES. 

AN  ANNUITY  is  a  yearly  income  arising  from  money, 
&c.  and  is  either  paid  for  a  term  of  years,  or  upon  a 
life.  Annuities  or  pensions  are  said  to  be  in  arrears, 
when  they  are  payable  or  due  either  yearly,  half-yearly, 
or  quarterly,  and  yet  remain  unpaid  for  any  number  of 
payments. 

The  sum  of  all  the  annuities,  for  the  time  they  have 
been  forborne,  together  with  the  interest  due  upon  each, 
is  called  the  amount. 

If  an  annuity  be  bought  off,  or  paid  all  at  once  at  the 
beginning  of  the  first  year,  {he  price  which  is  paid  for  it, 
is  called  the  present  worth. 

CASE  I. 

To  find  the  amount  of  an  annuity  at  Simple  Interest. 

RULE. — 1.  Find  the  interest  of  the  given  annuity  for 
1  year;  and  then  for  2,  3,  <fcc.  years,  up  to  the  given 
time  less  1.  2.  Multiply  the  annuity  by  the  number  of 
years  given,  and  add  the  product  to  the  whole  interest, 
and  the  sum  will  be  the  amount  required. 

EXAMPLES. 

1.  If  250  dollars,  yearly  annuity,  be  forborne  7  years, 
what  will  it  amount  to  in  that  time,  allowing  simple  in- 
terest at  6  per  cent,  per  annum  ? 

1st.  Interest  of  $250,  at  6  per  cent,  for  1  year=$lo. 
2yrs.  =  f30.  3yrs.=^$45.  4yrs.=:$60.  5yrs.=f 75, and 
6yrs.=$90.  And  2d.  $250x7  =  $1750. 

Then,  15+30+45+60+75+90+1750=$2065  Ans, 


EQUATION    OF    PAYMENTS.  141 

2.  A  house  is  leased  for  7  years,  at  400  dollars  per  an- 
num ;  and  the  rent  is  unpaid  during  the  whole  term; 
what  sum  is  due  at  the  end  of  the  lease,  simple  interest 
being  allowed,  at  6  per  cent,  per  annum  1 

Ans.  0304  dollars. 

CASE  II. 

To  find  the  present  worth  of  an  annuity  at  Simple  Interest. 

RULE.  —  Find  the  present  worth  of  each  year  by  itself, 

discounting  from  the  time  it    falls  due  ;  the   sum  pf  all 

these  present  worths,  will  be  the  present  worth  required. 

NOTE.  —  The  divisions  are  continued  decimally.  —  But 
it  is  deemed  more  equitable  to  allow  compound  interest, 
in  purchasing  annuities. 


1.  What  is  the  present  worth  of  400  dollars  per  annum, 
to  continue  5  years,  at  6  per  cent,  per  annum  ? 

106  ")  f  377,35849  ^present  worth  of  1  yr. 

112  I  |357,14285=  3d  yr. 

118  V:  100  :  :    400  :«{  338,98305=  3d  yr. 

124  j  I  322,580ti4~  4th  yr. 

130  j  [307,6923  «  5th  yr. 

Ans.  §1703,75733  =  $1703,75cts.  7m. 

2.  What  is  a  salary  of  $300  per  annum,  to  continue 
5  years,  worth,  in  ready  money  at  5  per  cent,  per  ann.  ? 

Ans.  $1309,31cts.+ 

3.  What  is   ,£250  yearly  rent,  to  continue  6  years, 
worth  in  ready  money,  at  3  per  cent.  ? 

Ans.  £1380  7s.  10f  AD- 


EQUATION OF  PAYMENTS. 

EQUATION  OF  PAYMENTS  is  finding  a  time  to  pay  at 
once  several  debts  due  at  different  times,  so  that  no  loss 
shall  be  sustained  by  either  party. 


142  EXCHANGE. 

RULE. — Multiply  each  payment  by  the  time  at  which 
it  is  due  ;  then  divide  the  sum  of  the  product  by  the  sum 
of  the  payments,  and  the  quotient  will  be  the  time  re- 
quired. 


EXAMPLES. 

1.  A  owes  B  1900  dollars,  to  be  paid  as  follows,  viz. 
500  dollars  in  6  months,  600  dollars  in  7  months,  and  800 
dollars  in  10  months;  what  is  the  equated  time  to  pay 
the  \vhole  debt  1 

500  x  6=3000 

600  x  7=4200 

800x10=8000 


1900         )  15200(8  months,  Ans. 
15200 

2.  A  owes  B  240  dollars  to  be  paid  in  six  months ;  but 
in  1^  months  pays  him  60  dollars,  and  in  4£  months  after 
that  80  dollars  more  ;  how  much  longer  than  six  months 
should  A  in  equity  defer  the  rest  1       Ans.  2T7«y  months. 

3.  I  owe  6512  dollars,  to  be  paid  ^  in  3  months,  £  in  5 
months,  £  in  ten  months,  and  the  remainder  in  14  months  ; 
at  what  time  ought  the  whole  to  be  paid  ? 

Ans.  6£  months. 

4.  A  owes  60  dollars  to  be  paid  in  90  days,  75  dollars 
in  60  days,  and  50  dollars  in  30  days  ;  what  is  the  equat- 
ed time  for  the  whole  to  be  paid  1       Ans.  61T6TFdays.+ 


EXCHANGE. 

BY  this  rule  merchants  know  what  sum  of  money  ought 
to  be  received  in  one  country,  for  a  given  sum  of  differ- 
ent specie  paid  in  another,  according  to  a  given  course  of 
exchange. 

By  the  following  table  the  moneys  of  foreign  nations 
may  be  reduced  to  that  of  the  United  States. 


EXCHANGE.  143 

TABLE 

Showing  the  value  of  the  coins  or  moneys  of  account, 
of  foreign  nations,  estimated  in  Federal  Money,  agreea- 
bly to  a  law  of  the  United  States  in  most  instances. 

8  cts. 

Pound  Sterling  of  Great  Britain,  4   44 

Pound  Sterling  of  Ireland,  -  -    4    10 

Livre  of  France,  -  0    18£ 

Guilder,  or  Florin,  of  Holland,     -  0    39 

Marc  Banco  of  Hamburgh,     -  -    0   33^ 

Rix  Dollar  of  Denmark,  and  Germany,         1    00 
Rial  Plate  of  Spain,  0    10 

Milre  of  Portugal,  -         -         -      1    24 

Tale  of  China  and  Japan,  1    48 

Pagoda  of  India,  -  -  1    94 

Rupee  of  Bengal,  -  0    55£ 

Ruble  of  Russia,  -  1    00 

Testoon  of  Italy,          -  0   23f 

Pound  English  W.  I.  Islands,     -  3    16f 

Livre  French  W.  I.  Islands,  -      0 

Franc  of  France,  -  0 

Rix  Dollar  of  Sweden,  .  -      1 

Do.     do.     of  Austria,     -  0 

Do.     do.     of  Poland  and  Prussia,  -       0 

Medio  of  Morocco,  -  -  J 

Piaster  of  Arabia,  -        1    00 

Or  of  Persia,  -  -  -     1    48 


I.— Of  Great  Britain. 
EXAMPLES. 

1.  in  ^45  10s.  sterling,  how  many  dollars   and  cents  ? 
A  pound  sterling  being   =444  cents.       Therefore, 

as  £1  :  444cts.  :  :  45,5£  :  20202cts.=$202,02cts.  Ans. 

2.  In  $500  how  many  pounds  sterling? 

As  444cts.  :  £1  :  :  50000cts.  :  ,£112  12s.  3d.+  Ans. 

NOTE. — This  method  is  not  so  exact  as  that  given  in 
Reduction  of  Currencies. 


144  EXCHANGE. 

2.—  Of  Ireland. 

1.  In  ,£85  7s.  6d.  Irish  money,  how  many  cents  ? 

£1  Irish=410cts. 

£.     cts.  £.  cts.         $     cts. 

Therefore,  as  1  :  410    :  :    85,375    :    35,003  J=350,03£ 

2.  In  $176,30cts.,  how  many  pounds  Irish  ? 

As  410cts.  :  £1    :  :    17630cts.  :  <£43  Irish.  Ans. 


3.—  Of  France. 

Accounts  are  there  kept  in  Hvres,  sols,  and  deniers. 
f  12  deniers,  or  pence,  make  1  sol,  or  shilling. 
(  20  sols,  or  shillings  1  livre  or  pound. 

EXAMPLES. 

1.  In  360  livres,  12  sols,  how  many  dollars  and  cents  ? 

1  livre,  of  France  =  18£cts.  or  185mills. 
I.         m.          L  m. 

As  1    :    185  ::  360,6  :  667  11  =$66,7  lets.  1m.  Ans. 

2.  What  sum  of  French  money  is  equal  to  $94,52cts. 
8m.  t         m.       liv.         m.  liv.  so.  den. 

As  185  :    1  :  :  94528  :   510  19'  2+  Ans. 

4.   Of  Holland,  8fc. 

Accounts  are  kept  there  in   guilders,  stivers,   grotes, 
and  peningens. 

C    8  peningens  make  1  grote'or  penny. 
<     2  grotes  1  stiver. 

(  20  stivers  1  guilder  or  florin. 

1  guilder  =39ets.  =390  mills. 

EXAMPLE. 

In  250  guilders,  16  stivers,  how  much  Federal  Money  ? 

guil.     cts.  guiL  $.  d.  c.  m. 

As  1    :    39      :     :      250,8      :     97  8  1    2  Ans. 

m.        g.  m.  g. 

As  390  :  1       :     :       97812    :    250,8  Proof. 

5.  —  Of  Hamburgh,  in   Germany. 

Accounts  are  kept  in  Hamburgh,   in  marcs,  shillings 
and  fennings,  and  by  some  in  rix  dollars. 


EXCHANGE.  145 

/  12  fennings  make  1  shilling  lub, 
J  16  shillings      -       1  marc. 
(    3  marcs          -       1  rix  dollar. 

1  marc=33£cts.  or  one  third  of  a  dollar. 

RULE  1. — Divide  the  marcs  by  3,  and  the  quotient  will 
be  dollars. 

EXAMPLE. 

Bring  584  marcs,  12  shillings  to  Federal  Money. 

3)584,75 

$194,9lc.  6§m.  Ans. 

RULE  2. — When  the  given  sum  is  Federal  Money,  mul- 
tiply it  by  3,  and  the  product  will  be  marcs  and  decimal 
parts. 

EXAMPLE. 

In  $3l5,70cts.  how  much  Hamburgh,  &c.  money  ? 
315,70 
3 


947,10=947  marcs,  1  shil,  7,  2  fennings.  Ans. 

6.—  Of  Spain. 

In  Spain,  they  keep  their  accounts  in  piastres,  rials, 
and  maravedies. 

{34  naaravedies  of  plate  make  1  rial  of  plate. 
8  risfts  of  plate  make  1  piastre,  or  piece  of  8. 
1  rial =10  cents,  or  1  dime. 

RULE  1. — To  reduce  rials  to  Federal  Money,  point  off 
the  rigfit  hand  figure  of  the  given  sum,  and  as  it  stands, 
it  is  the  answer  in  dollars  and  dimes. 

EXAMPLE. 

In  968  rials,  how  many  dollars,  &c.  ? 

Ans.  96,8  ^=$96,8  dimes,  or  $96,  SOcts- 

RULE  2. — To  reduce  cents  into  rials,  divide  the  cents 
by  10. 

N 


146  EXCHANGE. 

EXAMPLE. 
In  $94,65cts.  bow  many  rials,  &c.  ? 

9465— 10  =946,5  =946  rials,  17  maravedies.  Ans. 

7.— Of  Portugal 

In  Portugal,  accounts  are  kept  in  niilrez  and  rez,  reck- 
oning 1000  rez  to  a  milre,  as  its  name  imports. 
1  milre=z:l24  cents. 

RULE  1. — To  reduce  milrez  into  Federal  Money,  mul- 
tiply the  given  sum  by  124,  and  the  product  will  be  cents  ; 
if  the  given  sum  contain  rez,  multiply  as  before,  and  cut 
off  3  figures  from  the  right  of  the  product ;  the  answer 
will  be  cents,  and  decimals  of  a  cent.  If  the  rez  are  less 
than  100,  prefix  a  cipher  to  them,  before  you  annex  them 
to  the  milrez.  A  comma,  or  decimal  point,  is  used  to 
separate  rez  and  milrez. 

EXAMPLES. 

1.  In  580  milrez  how  many  cents  ? 

580x124=71920  cents  =  $719,20cts.  Ans. 

2.  In  125  milrez,  96  rez,  how  many  cents  ? 
125,096 x  124  =  1551  l,904cts.  or  $l55,llcts.  9m.+Ans. 

RULE  2. — To  reduce  cents  into  milrez,  divide  the  sum 
by  124,  and  if  decimals  arise,  continue  the  division  to 
three  places  of  decimals  ;  the  whole  numbers  in  the  quo- 
tient will  be  milrez,  and  the  decimals  will  be  rez. 

EXAMPLES. 

1.  How  many  milrez  in  6878  cents  ? 

6878—124=55,467+  or  55milr.  467  rez.    Ans. 

2.  In  $31,09cts3  mills,  how  many  milrez  of  Portugal  ? 

c.  m. 
3109,3-;-124=«25,075=25milr.  75rez.  Ans. 

8. — Of  East-India  Money. 

RULE. — To  reduce  India  Money  to  Federal.  Multiply 
tales  of  China  by  148;  pagodas  of  India  by  194;  and 
rupees  of  Bengal,  by  55J  :  and  the  product,  in  every  case, 
will  be  cents.  To  change  Federal  Money  into  the  before- 


VULGAR    FRACTIONS.  147 

named  moneys,  divide  the  given  sum  by  one  of  the  above 
numbers,  as  the  case  may  be. 

EXAMPLES. 

1.  In  1000  tales  of  China,  how  many  dollars  ? 

Ans.  $1480. 

2.  In  1420  tales  of  China,  how  many  dollars,  &,c.  ? 

Ans.  §2101, 6Gcts. 

3.  In  $948,68cts.  how  many  tales  of  China  ? 

Ans.  641. 

4.  In  482  pagodas  of  India,  how  many  dollars  and 
cents  1  Ans.  8935,8cts. 

5.  In  S135,80cts.  how  many  pagodas  of  India? 

Ans.  70. 

6.  In  920  rupees  of  Bengal,  how  much  Federal  Money  ? 

Ans.  $510,60cts. 

7.  In  $54,39cts.  how  many  rupees  of  Bengal  ? 

Ans.  98. 

A  compound  Example  in  European  Money. 

Suppose  a  merchant  of  Hallowell  has  effects  at  London, 
to  the  amount  of  $5460,  which  he  can  remit  by  way  of 
Lisbon,  at  1  milre  per  dollar,  and  thence  to  Boston,  at 
$l,50cts.  per  milre  ;  or  by  way  of  Paris,  at  5  livres  per 
dollar  ;  thence  to  Brest,  at  6  livres  per  crown,  and  thence 
to  Washington,  at  $I,25cts.  per  crown;  which  will  be 
the  most  advantageous  way  of  remitting  ? 

A        f  Remitting  by  way  of  Lisbon  is  most 
3*  \      advantageous  by  $2502,50cts. 


VULGAR  FRACTIONS. 

VULGAR  FRACTIONS  were  briefly  introduced  immediate- 
ly after  Duodecimals  ;  and  some  general  definitions,  and 
a  few  such  problems  as  were  necessary  to  prepare  the 
scholar  for,  and  to  lead  him  into  decimals, were  there  given. 
Those  general  definitions  he  is  requested  to  read  anew. 


148  VULGAR    FRACTIONS. 

Vulgar  Fractions  are  either  proper,  improper,  single, 
compound  or  mixed. 

1.  A  proper  fraction  is  when  the  numerator  is  less 
than  the  denominator  :  as,  f  ,  f,  f,  &,c.  which  mean  £  of 
1,  f  of  l,f  of  l,&c. 

2.  An  improper  fraction  is  when  the  numerator  ex- 
ceeds the  denominator  :  as,  f  ,  -yT°-,  &c. 

3.  A  single  fraction  is  a  simple  expression  for  any 
number  of  parts  of  the  integer. 

4.  A  compound  fraction  is  the  fraction  of  a  fraction  : 
as,  £  of  f,  |  of  £,  &c.,  which  mean  £  of  f  of  l,f  of  £  of 
1,  &c. 

5.  A  mixed  number  is  composed  of  a   whole   number 
and  a  fraction  ;  as,  8-J,  12T97j,  &,c. 


NOTE.  —  Any  number  may  be  expressed  like  a  fraction 
by  writing  1  under  it  :  Thus  -f-  means  6  ones  or  6. 

A  fraction  having  a  fraction  or  mixed  number  for  its 
numerator  or  denominator,  or  both,  is  called  a  complex 
fraction.  A  fraction  denotes  division,  and  its  value  is 
equal  to  the  quotient  obtained  by  dividing  the  numerator 
by  the  denominator  :  thus  -^  is  equal  to  3,  and  ^§-  equal 
to  4.  Therefore,  if  the  numerator  be  less  than  the  de- 
nominator, the  value  of  the  fraction  is  less  than  1.  If 
the  numerator  be  the  same  as  the  denominator,  the  frac- 
tion is  just  equal  to  1.  And  if  the  numerator  be  greater 
than  the  denominator,  the  fraction  is  greater  than  1. 

6.  The  common  measure   of  two   or   more  numbers  is 
that  number  which  will  divide  each  of  them  without  a 
remainder  ;  and  the  greatest  number  that  will  do  this,  is 
called  the  greatest  common  measure. 

7.  A  number  which  can  be  measured  by  two  or  more 
numbers,  is  called  their  common  multiple  ;  and  if  it  be  the 
least  number,  which  can  be  so  measured,  it  is  called  their 
least  common  multiple. 

PROBLEM  1. 

To  find   the  greatest  common   measure    of  two    or   more 
numbers. 

RULE.—  1.  If  there  be  two  numbers  only,  divide  the 
greater  by  the  less,  and  this  divisor  by  the  remainder,  ancl 


VULGAR    FRACTIONS.  149 

so  on  ;  always  dividing  the  last  divisor  by  the  last  remain- 
der, until  nothing  remains  ;  then  will  the  last  divisor  be 
the  greatest  common  measure  required. 

2.  When  there  are  more  than  two  numbers,  find  the 
greatest  common  measure  of  two  of  them,  as  before  ;  and 
next  find  the  greatest  common  measure  of  that  common 
measure    and   one  of  the   other  numbers  ;    and  so   on, 
through  all  the  numbers  to  the  last ;  then  will  the  greatest 
common  measure  last  found  be  the  answer. 

3.  If  one  happen  to  be  the  common  measure,  the  given 
numbers  are  prime  to  each  other,  and  found  to  be  incom- 
mensurable. 

EXAMPLES. 

1.  Required  the  greatest  common  measure  of  918,1998 
and  522. 

918)1998(2 

1836  So  54  is  the  greatest  common 

measure  of' 1998  and  918— 
162)918(5  Hence  54)522(9 

810  486 

108)162(1  36)54(1 

108  36 

54)108(2  18)36(2 

108  36 

Therefore  18  is  the  answer  required. 

2.  What  is  the  greatest  common  measure  of  612  and 
540?  Ans.  36. 

3.  What  is  the  greatest  common  measure  of  117  and 
91  1  Ans.  13. 

PROBLEM  2. 

To  find  the  least  common  multiple  of  two  or  more  numbers. 

RULE. — 1.  Divide  by  any  number,  that  will  divide  two 

or  more  of  the  given  numbers  without  a  remainder,  and 

set  the  quotient,  together   with  the   undivided   numbers, 

in  a  line  beneath. 

2.  Divide  the  second  line  as  before,  and   so  on,   until 
there  are  no  two  numbers  that  can  be  divided  ;  then  the 
N  2 


150  VULGAR   FRACTIONS. 

continued  product  of  the  divisors  and  quotient  will  give 
the  multiple  required. 


EXAMPLES. 

1.  What  is  the  least  common  multiple  of  3,  5,  8,  and 
101 

2)3  5  8  10 

5)3  545 

3141     Then  2x5x3x4=120  Ans. 

2.  What  is  the  least  common  multiple  of  9,  8,  15,  16  ? 

Ans.  720. 

3.  What  is  the  least  number  that  3,  4,  8,  and  12  will 
measure  ?  Ans.  24. 

4.  What  is  the  least  number  that  can  be  divided  by  the 
9  digits  without  a  remainder  Ans.  2520. 


REDUCTION  OF  VULGAR  FRACTIONS. 

REDUCTION  OF  VULGAR  FRACTIONS  is  the  bringing 
them  out  of  one  form  into  another,  in  order  to  prepare 
them  for  the  operations  of  addition,  subtraction,  &c. 


CASE  I. 

To  abbreviate  or  reduce  fractions  to  their  lowest  terms. 

RULE. — Divide  the  terms  of  the  given  fraction  by  any 
number  that  will  divide  them  without  a  remainder,  &c. 
as  in  Rule  of  Problem  1,  page  76.  Or  divide  both  the 
terms  of  the  fraction  by  their  greatest  common  measure, 
and  the  quotients,  will  be  the  terms  of  the  fraction  re- 
quired. If  a  fraction  have  ciphers  on  the  right  hand  of 
both  its  terms,  it  may  be  reduced  by  cutting  off  an  equal 
number  from  both. 


VULGAR   FRACTIONS.  151 

EXAMPLES. 

1.  Reduce  JM  to  its  lowest  terms. 

(4)     (3)     (4) 

it*  =  lf=iS=f  the  answer. 
Or  thus,  144)240(1 
144 

96)144(1 
96 

Therefore  48  is  the  greatest 
48)96(2         common  measure,  and 
96  48)^=f  Ans. 

NOTE. — 1.  Any  number  ending  with  an  even  number, 
or  a  cipher,  is  divisible  by  2. 

2.  Any  number  ending  with  5  or  0,  is  divisible  by  5. 

3.  If  the  right  hand  place  of  any  number  be  0,  the 
whole  is  divisible  by  10. 

4.  If  the  two  right  hand  figures,  of   any  number  are 
divisible  by  4,  the  whole  is  divisible  by  4. 

5.  If  the  sum  of  the  digits,  constituting  any  number, 
be  divisible  by  3  or  9,  the  whole  is  divisible  by  3  or  9. 

6.  All  prime  numbers,  except  2  and  5,  have  1,  3,  7,  or 
9,  in  the  place  of  units;  and  consequently  all  other  num- 
bers are  composite,  and  capable  of  being  divided. 

7.  When  numbers  with  the  sign  of  addition  or  subtrac- 
tion between  them,  are  to  be  divided  by  any  number, 
each  of  the  numbers  must  be  divided.     Thus, 

4+8+10 

=2+4+5=11 

2 

8.  But  if  the  numbers  have  the  sign  of  multiplication 
between  them,  only  one  of  them  must  be  divided.     Thus, 

3x8x10  x3x4xlO  1x4x10  1x2x10 

, m  2_0 fffl 

2x6  1x6  1x2  1x1 

9.  If  both  the  numerator  and  denominator  of  a  frac- 
tion be  multiplied  or  divided  by  the  same   number,  the 
fraction  will  still  retain  its  original  value. 

Let  £  and  T9^  be  two  fractions  proposed  ;  then  1x1=^; 
and  T9^-rf=f'     That  is,  if  the  numerator  4,  and  denom- 


152  VUGLAR   FRACTIONS. 

inator  5,  of  the  first  fraction,  be  each  multiplied  by  the 
same  number  2,  the  produced  fraction  T8^  is  equal  to  the 
proposed  one  f.  For  the  numerator  and  denominator  of 
the  produced  fraction,  are  in  the  same  proportion  as  the 
numerator  and  denominator  of  the  proposed  one.  Also, 
if  the  numerator  9,  and  the  denominator  12,  of  the  second 
fraction,  be  each  divided  by  the  same  number  3,  the  frac- 
tions J  and  T92-  are  equal  for  the  same  reason. 


2.  Reduce  -£fe  to  its  lowest  terms.  Ans,  -j^-. 

3.  Reduce  -f/-^  to  its  lowest  terms.  Ans.  £f. 

4.  Reduce  T4T9g^  to  its  lowest  terms.  Ans.  fff  . 

5.  Reduce  ff-Jf  to  its  lowest  terms.  Ans.  •££. 

CASE  II. 

To  reduce   a   mixed  number   to    its    equivalent   improper 
fraction. 

RULE.  —  Multiply  the  whole  number  by  the  denomina- 
tor of  the  fraction,  and  add  the  numerator  to  the  pro- 
duct; then  that  sum  written  above  the  denominator, 
will  form  the  fraction  required. 

EXAMPLES. 

1.  Reduce  27|-  to  its  equivalent  improper  fraction. 

27 
9 

243 
2 


245     Or  27x9 +2 

=-|^  the  answer. 

9  9 

2  Reduce  514T^  to  an  improper  fraction.  Ans.  ^ 

3.  Reduce  I2|f  to  an  improper  fraction.    Ans. 

4.  Reduce  79{-f  to  an  improper  fraction.  Ans. 

5.  Reduce  100||  to  an  improper  fraction.  Ans. 


TULfSAR   FRACTIONS.  153 

CASE  III. 

To  reduce  an  improper  fraction  to  its  equivalent  whole  or 

mixed  number. 

RULE.  —  Divide  the  numerator  by  the  denominator,  and 
the  quotient  will  be  the  whole  or  mixed  number  required. 

EXAMPLES. 
1.  Reduce  &££•  to  its  equivalent  whole  or  mixed  Dumber. 


19)081(61  A 

90 

21 
16 

5  or  ^=981—16=61^  the  answer. 

2.  Reduce  5ry9-  to  its  equivalent  whole  or  mixed  num- 
ber. Ans.  12ff  . 

3.  Reduce  *£g-  to  its  equivalent  whole  or  mixed  num- 
ber. Ans.  2f. 

4.  Reduce  -^  to  its  equivalent  whole  or  mixed  num- 
ber. Ans.  7. 

5.  Reduce  **£$£*•  to  its  proper  terms.  Ans.  1209|f  J. 

CASE  IV. 

To  reduce  a  whole  number  to  an  equivalent  fraction,  hav- 

ing a  given  denominator* 

RULE.  —  Multiply  the  whole  number  by  the  given  de- 
nominator, and  place  the  product  over  the  said  denomi- 
nator, and  it  will  form  the  fraction  required. 

EXAMPLES. 

1.  Reduce  7  to  a  fraction,  whose  denominator  shall  be 
9.  7x9=63,  and  &£  Answer. 

And  ^.=63-5-9=7  Proof. 

2.  Reduce  13  to  a  fraction,   whose  denominator  shall 
be  12.  Ans.  J^/. 

3.  Reduce  746  to  a  fraction,  whose  denominator  shall 
be  60,  Ans. 


154  VULGAR    FRACTIONS. 

CASE  V. 

To  reduce  a  compound  fraction  to  an  equivalent  single  one. 

RULE. — Multiply  all  the  numerators  together  for  the 
numerator,  and  all  the  denominators  together  for  the  de- 
nominator, and  they  will  form  the  fraction  required. 

If  part  of  the  compound  fraction  be  a  whole  or  mixed 
number,  it  must  be  reduced  to  an  improper  fraction  by 
one  of  the  former  cases. 

When  it  can  be  done,  divide  any  two  terms  of  the 
fraction  by  the  same  number,  and  use  the  quotients  in- 
stead thereof. 

EXAMPLES. 

1.  Reduce  §  of  £  of  f-  of  T8T  to  a  single  fraction. 
2x3x4x8 

=|-ff  =|f  the  answer. — Or,  by  expunging 

3x4x5x11 

equal  numerators  and  equal  denominators,  the  answer 
will  be  as  before,  =^f. 

2.  Reduce  f  of  f  of  f  to  a  single  fraction.     Ans.  •%. 

3.  Reduce  T^  of  T77j  of  T8^  of  10  to  a  single  fraction. 

Ans.  4/^-f- . 

4.  Reduce  J-  of  f  of  f  to  a  single  fraction.  Ans.  8^8T. 

CASE  VI. 

To  reduce  fractions  of  different  denominators  to  equiva- 
lent fractions,  having  a  common  denominator. 

RULE. — Multiply  each  numerator  into  all  the  denomi- 
nators except  its  own,  for  a  new  numerator;  and  all  the 
denominators  continually  for  the  common  denominator; 
first  reducing  the  fractions  to  their  lowest  terms,  &c. 

NOTE. — By  Note  9,  in  case  1,  it  will  be  seen,  that  sev- 
eral fractions  of  different  denominators  may  be  readily 
reduced  to  a  common  denominator.  Thus  -J-  may  be  re- 
duced to  the  same  denominator  as  f  by  multiplying  its 
terms  by  3,  by  which  it  becomes  f-.  Also  ^-,  f ,  and  ^ 
may  be  reduced  to  a  common  denominator,  by  multiply- 
ing the  terms  of  the  first  fraction  by  6,  of  the  second  by 
3,  and  dividing  those  of  the  last  by  5.  And  so  of  others. 


VULGAR    FRACTIONS.  155 

EXAMPLES. 

1.  Reduce  £,  f ,  and  f-  to  equivalent  fractions,  having  a 
common  denominator. 

1  x5x7=35  the  new  numerator  for  i. 
3x2x7=42        do.  do.  f. 

4x2x5=40         do.  do.  f 

2x5x7=70  the  common  denominator. 
Therefore  the  equivalent  fractions  are  f£,  tf »  an(*  tS- 
the  answer. 

2.  Reduce  T6<y,  f,  £,  and  f  to  equivalent  fractions,  hav- 
ing a  common  denominator.  Ans.  f  |f ,  f~J$,  ^$7,  and  -| f  g. 

3.  Reduce  £,  f  of  f ,  5^  and  T2^  to  a  common  denomi- 
nator. Ans.  #$,  ff§,  Jy^,  ^6TV 

4.  Reduce  |^,  f  of  1^,  T9T  and  f  to  a  common  denom- 
inator. Ans.  iM 

CASE  VII. 

To  find  the   value  of  a  fraction  in  any  known  parts  of 

the  integer. 

RULE. — Multiply  the  numerator  by  the  parts  in  the 
next  inferiour  denomination,  and  divide  the  product  by 
the  denominator  ;  and  if  any  thing  remain,  multiply  it 
by  the  next  inferiour  denominator,  and  divide  by  the  de- 
nominator as  before  ;  and  so  on,  as  far  as  necessary  ;  and 
the  quotients  placed  after  one  another,  in  their  order,  will 
be  the  answer  required. 

EXAMPLES. 

1.  What  is  the  value  of  f  of  a  shilling  ?     Ans. 
3 
12 

8)36(4d. 
32 

4 
4 

8)16(2qrs. 
16 


156  VULGAR   FRACTIONS. 

2.  What  is  the  value  of  T5^  of  a  dollar  1 

Ans.  41cts.  6f  mills. 

3.  What  is  the  value  of  f  of  a  mile  ? 

Ans.  4fur.  22pol.  4yds.  2|feet. 

4.  What  is  the  value  of  f  of  a  month  ? 

Ans.  3w.  Id.  9h.  36m. 

5.  What  is  the  value  of  T7F  of  an  acre  ? 

Ans.  Irood,  SOpoles. 

6.  What  is  the  value  of  f  of  f  of  £  of  $49,95cts.  1 

Ans.  $5,55cts. 

CASE  VIII. 

To  reduce  a  fraction   of  one   denomination  to  that  of 

another,  retaining  the  same  value. 

RULE.  —  Make  a  compound  fraction  of  it,  and  reduce 
it  to  a  single  one. 

EXAMPLES. 

1.  Reduce  f  of  a  penny  to  the  fraction  of  a  pound. 

£  of  j±  of  2V=TA^=^F  the  answer. 
And  s-fg-  of  ^  of  -y=f|f=|d.  the  proof. 

2.  Reduce  ^  of  a  farthing  to  the  fraction  of  a  pound 

Ans.  Tsfoj. 

3.  Reduce  -     of  anill  to  the  fraction  of  a  dollar. 


4.  Reduce  -^£  to  the  fraction  of  a  penny. 

Ans.  -4_°-= 

5.  Reduce  f  of  a  pound  Avoirdupois  to  the  fraction  of 
an  cwt.  Ans.  Tf¥=3f?. 

6.  Reduce  T%  of  a  month  to  the  fraction   of  a  day. 

Ans.  ff 

7.  Reduce  7s.  3d.  to  the  fraction  of  a  pound. 

5.  d.  s. 

73  20  in  a  £ 

12  12 

87  numerator.  240  denominator. 


8.  Reduce  6  furlongs,  16  poles  to  the  fraction  of  a  mile. 

Ans.  f. 


VULGAR    FRACTIONS.  157 

ADDITION  OF  VULGAR  FRACTIONS. 

RULE. — Reduce  compound  fractions  to  single  ones  ; 
mixed  numbers  to  improper  fractions  ;  fractions  of  differ- 
ent integers  to  those  of  the  same  ;  and  all  of  them  to  a 
common  denominator  ;  then  the  sum  of  the  numerators, 
written  over  the  common  denominator,  will  be  the  sum 
of  the  fractions  required. 

NOTE  1. — In  adding  mixed  numbers  that  are  not  com- 
pounded with  other  fractions,  find  first  the  sum  of  the 
fractions,  to  which  add  the  whole  numbers  of  the  given 
mixed  number. 

NOTE  2. — When  adding  fractions  of  money,weight,&c. 
reduce  fractions  of  different  integers  to  those  of  the  same 
integer.  Or,  find  the  value  of  each  fraction  by  Case  7, 
in  Reduction,  and  then  add  them  in  their  proper  terms. 

EXAMPLES. 

1.  Add  3f ,  I,  f  of  I  and  7  together. 

First,  3f=^-,  f  of  £=f  §=T^,  7HF- 
Then  the  fractions  are  5g9-,  -|,  T7<y  and  £. 
29x8  xiOx  1=2320 
7x8xlOx   1=  560 
7x8x  8x   1=  448 
7x8x  8x10=4480 


7808 

=12|ff=12 

8x8x10x1=640 

2.  Add  f ,  7£,  and  -J  of  f  together.  Ans.  S| . 

3.  What  is  the  sum  of  £  of  95  and  J  of  14  ? 

Ans.  43-|4. 

4.  Add  19,  7,  and  J-  of  f  together.  Ans.  26|. 

5.  Add  f  and  17^  together.  Ans.  18 J. 

6.  What  is  the  sum  of  £}9  f  s.  and  T5?  of  a  penny  ? 

Ans.  fi$fs.  or~3s.  Id.  l^qrs. 

7.  Add  $  of  15£.  £3f ,  £  of  f  of  f  of  a  pound  and  f 
of  f  of  a  shilling  together.  Ans.  £7  1 7s.  5fd. 

8.  Add  f  of  a  yard,  £  of  a  foot  and  f  of  a  mile  to- 
gether. Ans.  660yds.  2ft.  9in. 

9.  Add  £  of  a  week,  ^  of  a  day  and  £  of  an  hour  to- 
gether. Ans.  2days,  14£hours. 

O 


158  VULGAR   FRACTIONS. 

10.  Add  f-  of  a  ton  to  •£$  of  an  hundred  weight, 

a  qr.  Ans.  12cwt.  Iqr.  7ft.  13oz.  llfdrs. 

11.  Suppose  I  have  f  of  a  ship  worth  $6000,  and  that 
I  buy  another  person's  share  of  her,  which  is  T\  ;  what 
part  of  her  belongs  to  me  then,  and  what  is  it  worth  1 

Ans.  I  have  ft,  and  it  is  worth  $4125. 

SUBTRACTION    OF  VULGAR    FRACTIONS. 

RULE. — Prepare  the  fractions  as  in  addition,  and  the 
difference  of  the  numerators  written  above  the  common 
denominator,  will  give  the  difference  of  the  fractions  re- 
quired. 

NOTE  1. — In  subtracting  mixed  numbers,  when  the 
fraction  in  the  subtrahend  is  greater  than  that  in  the  min- 
uend, subtract  the  numerator  of  the  subtrahend  from  the 
denominator,  and  to  the  difference  add  the  numerator  of 
the  minuend,  and  carry  one  to  the  integer  in  the  subtra- 
hend. If  the  minuend  contain  no  fraction,  proceed  in 
the  same  way,  there  being  then  nothing  to  add  to  the 
difference. 

NOTE  2. — In  fractions  of  money,  weights,  &c.  you 
may  find  the  value  of  each  of  the  given  fractions,  by  Case 
?,  in  Reduction,  and  then  subtract  them  in  their  proper 
terms. 

EXAMPLES. 

1.  From  f  take  f  of  f 

|  of  f=/T 
££ — /T=£f  =4-  the  answer. 

2.  From  ||J  take  £.  Ans.  -f-f?. 

3.  From  7ll  take  ft.  Ans.  70ff. 

4.  From  <££  take  f  of  a  shilling.  Ans.  16s.  9d. 

5.  From  foz.  take  £  of  a  pennyweight. 

Ans.  lldwt.  3grs. 

6.  From  £cwt.  take  £F  of  a  pound. 

Ans.  Iqr.  27  tb-  6oz.  lOfdrs. 

7.  From  7  weeks  take  9  days  T7^. 

Ans.  5w.  4d.  7h.  12m. 

8.  From  4  days  7£  hours  take  1  day  9  hours  ^. 

Ans.  2d.  22h.  20m. 


VULGAR    FRACTIONS.  159 

9,  Suppose  I  own  f  of  a  farm,  which  is  worth  $3600, 
and  that  I  sell  f  of  my  share  ;  what  part  of  it  have  I  left 
and  what  is  it  worth  ] 

Ans.  £fc  ;  and  worth  $750. 

MULTIPLICATION  OF  VULGAR  FRACTIONS. 

RULE.  —  Reduce  compound  fractions  to  single  ones, 
mixed  numbers  to  improper  fractions,  and  those  of  differ- 
ent integers  to  the  same  ;  then  multiply  all  the  numera- 
tors together  for  the  numerator,  and  multiply  all  the  de- 
nominators together  for  the  denominator,  of  the  product 
required.  But  always,  before  multiplying,  cancel  equal 
numerators  and  denominators,  and  divide  those  that  are 
divisible  by  the  same  numbers,  both  here  and  in  division, 
agreeably  to  what  is  seen  in  Note  8  of  Case  1,  in  Reduc- 
tion, page  150,  and  in  Case  5,  page  154. 

NOTE.  —  A  fraction  is  best  multiplied  by  an  integer,  by 
dividing  the  denominator  by  it,  if  possible,  but  if  that 
cannot  be  done,  multiply  the  numerator  by  it. 

EXAMPLES. 

1.  Required  the  continued  product  of  2^,  -J,  £,  of  f 
and  2.  1x5 

2JHb  *  of  f  =  -  =T\»  and  2=f  . 

3x6 

^5x1x5x2 

Then  J  X£XT5FXf  =  --  =T^  Ans. 
2x8x18x1 

2.  Multiply  f  by  T3T.  Ans.  TV 

3.  Multiply  4£  by  \.  Ans.  TV 

4.  Multiply  f  of  8,  by  £  of  5.  Ans.  21. 

5.  Multiply  7£  by  9f  Ans.  69f  . 

6.  Multiply  4J-,  |-  of  ^,  and  184  continually  together. 

Ans.  9T|^. 

7.  What  is  the  product  of  5,  f  ,  f  ,  of  f  ,  and  4£  ? 

Ans. 


DIVISION  OF  VULGAR  FRACTIONS. 

RULE.  —  Prepare  the  fractions  as  in  Multiplication  ; 
then  invert  the  divisor,  and  proceed  as  in  Multiplication. 
The  product  will  be  the  quotient  required. 


160    RULE  OF  THREE  IN  VULGAR  FRACTIONS. 

NOTE.  —  A  fraction  is  divided  by  an  integer,  by  dividing 
the  numerator  by  it,  if  possible,  but  if  it  will  not  exactly 
divide,  then  multiply  the  denominator  by  it. 

EXAMPLES. 

1.  Divide  |  of  19  by  f  of  £  . 

1x19      19 
|  of  19=—  --  =—  ,  and  f  of  f  =T62-=£  ; 

5x  1       5 
19x2 

=-^-=7f  the  quotient  required. 


5x1 

2.  Divide  ^  by  f.  Ans.  Iff. 

3.  Divide  99  by  108.  Ans.  A9ff=ii- 

4.  Divide  f  of  f  by  £  Of  f  .  Ans.  l£. 

5.  Divide  3£  by  9£.  Ans.  £. 

6.  Divide  £  by  4.  Ans.  ^ 

RULE  OF  THREE  IN  VULGAR  FRACTIONS. 

RULE.  —  Prepare  the  fractions  as  in  Multiplication  ; 
and  state  the  question  as  directed  in  the  Rule  of  Three 
in  whole  numbers  ;  then  invert  the  first  term  in  the  stat- 
ing, and  multiply  all  the  three  terms  continually  together, 
numerators  by  numerators  and  denominators  by  denomi- 
nators, and  the  product  will  be  the  answer  in  the  same 
name  as  the  second  or  middle  term. 

EXAMPLES. 

1.  If  f  of  a  yard  cost  T7^  of  a  pound,  what  will  T\  of 
an  ell  English  cost  1 

3x4x1 
First  f  of  a  yard=f  of  f  of  £=  ---  =£f  of  an  el!- 

5x1x5 

Ell        £.        Ell. 
Then  £§    :    &    :  :    A  - 
And  f|-XTVxT\=f|-^=9s.  8fd.  Answer. 

NOTE.  —  Here,  25  and  15  are  divisible  by  5,  giving  5 
and  3  for  quotients  ;  and  6  and  one  of  the  12s.  are  divisi- 
ble by  6,  giving  1  and  2  for  quotients  ;  then  the  numerators, 
are  5,  7,  and  1,  whose  product  is  05  ;  and  the  denomina- 
tors 12,  2,  and  3,  whose  product  is  72.  And  thus  always 


RULE    OF    THREE    IN   DECIMALS.  161 

proceed,  in  multiplying  vulgar  fractions,  by  cancelling 
equal  numerators  and  denominators,  or  dividing  those 
that  are  divisible  by  the  same  number. 

2.  If  f  of  a  yard  cost  £  of  a  dollar,  what  will  40f  yds. 
come  to?  Ans.  $59,42cts.  7T12-m. 

3.  If  50f  bushels  of  wheat  cost  $59^,  what  is  it  per 
bushel?  Ans.  $  1,1  Gets.  6fm. 

4.  If  I  buy  lOOffo.  of  butter  for  $10£,  how  many  hun- 
dred  can  I  have  for    $105fi?         Ans.  10^  hundred. 

5.  If  ^  of  £  of  a  yard  of  lace  cost  30cts.,  how  much 
is  it  per  yard  ?  Ans.  $2,40cts. 

6.  If  f  of  a  gallon  of  wine  cost  f  of  a  dollar,  what  will 
£  of  a  tun  cost  ?  Ans.  $140. 

7.  A  younger  brother  received  $2200,  which  was  just 
y5^  of  his  elder  brother's  fortune  ;  and  3£  times  the  el- 
der's money  was  j-  as  much  again  as  the  father    was 
worth  ;  what  was  the  father  worth  ?  Ans.  $11000. 

8.  When  the   days  are   13f   hours  long,    a    traveller 
performs  his  journey  in  35£  days ;  in   how  many  days 
will  he  perform  the  same  journey,  when  the  days  are 
11T9Q-  hours  long?  Ans.  40f^f  days. 

9.  A  schoolmaster  being  asked  how   many  scholars  he 
had,  answered,  If  I  had  as  many,  and^-  as  many,  and  % 
as  many,  I  should  have  99 ; — how  many  had  he  ?    Ans.  36. 


RULE  OF  THREE  IN  DECIMALS. 

RULE. — Reduce  vulgar  fractions  to  decimals,  and  com- 
pound numbers  either  to  decimals  of  the  higher  names, 
or  to  integers  of  the  lower,  as  also  the  first  and  third  terms 
to  the  same  name;  then  state  the  question  and  proceed 
as  in  whole  numbers ;  the  fourth  term  will  be  the  answer. 

EXAMPLES. 

1.  If  f  of  a  yard  cost  f  of  a  pound,  what  will  £  of  an 
ell  English  cost? 

f=r,375yd.  f =,4£.     ±  ell=TVyd.-,3125yd. 

Yd.        £.         Yd.         £. 

As  ,375  :  ,4  :  :  ,3125  :  ,333+  =6s.  8d.     Ans. 
02 


162    DOUBLE    RULE    OF    THREE    IN   VULGAR   TRACTIONS. 

2.  If  Icwt.  Iqr.  16f  fc.  of  sugar  cost  $10,9cts.,  what 
will  9cwt.  3qr.  cost,  at  the  same  rate  ? 

As  1,4  :  10,09:  :  9,75  :  70,269  +  =$70,26cts.9m.+  Ans. 

3.  A  man  bought  5,8  tuns  of  oil,  for  $266,  but  by  ac- 
cident, lost  50,9  gallons  ;  how  must  he  sell  the   residue 
per  gallon,  so  as  not  to  lose  in  his  bargain  1 

Ans.  I8cts.  8m.  + 

4.  If  12  oxen  graze  down  16,25  acres  of  grass   in  20 
days,  how  much  of  the  same  pasture  will  suffice  24  oxen 
100  days  1  Ans.  162,5  acres. 

5.  By  how  many  men  would  417,6  acres  be  mowed  in 
12  days,  if  5  men  mow  -J-  of  that  quantity  in  £  the  time  ? 

Ans.  20  men. 

6.  Two  men  bartered  ;  A  had  24,07  yards  of  linen,  for 
which  B  gave  him  25,6  ells  of  Holland,  at  45cts.  per  ell ; 
what  was  the  linen  per  yard  ?  Ans.  47cts.  8m. + 

7.  A  man  bought  a  piece  of  broadcloth  for  $S5,20cts., 
at  the  rate  of  $4,26cts.  per  yard  ;  how  many  yards  did 
it  contain  1  Ans.  20yds. 

8.  If,  during  tbe  tide   of  ebb,   a  boat  should  set  out 
from  Hallowell  for  Bath,  and  at  the  same  instant  another 
should  put  off  at  Bath  for  Hallowell,  taking  the  distance 
by  water  at  25  miles  ;  the  tide  forwards  one,  and  retards 
the  other,  say  2J-  miles  an  hour;  the  boats  are   equal  in 
size  and  form,  and  equally  laden,  the  rowers  equally  good, 
and  in  the  common  way  of  working,  in  still  water,  would 
proceed  at  the  rate  of  5  miles  an  hour  ;  where  in  the  riv- 
er would  the  two  boats  meet  ? 

Ans.  They  will  meet  18f  miles  from  Hallowell. 

9.  Bought  12  pieces  of  Holland  for  £404£$,  at   $1^ 
per  ell  Flemish  ;  how  many  yards  were  in  each  piece  ? 

Ans.  22|yds. 

DOUBLE   RULE   OF   THREE   IN    VULGAR 
FRACTIONS. 

RULE. — State  the  question  as  in  the  same  rule  in  whole 
numbers;  invert  the  fractions  which  stand  in  and  under 
the  first  term  ;  then  proceed  as  in  the  Rule  of  Three  in 
Vulgar  Fractions* 


SIMPLE    INTEREST    BY  DECIMALS. 


163 


EXAMPLES. 

I.  If  £2  10s.  be  the  wages  of  15  men  for  6  days,  what 
will  be  the  wages  of  12  men  for  18£  days  1 

J^  men  )    .   5^    ,         (  -^  men 
f  days  J    :  **  I  Af  days    ' 

==6^  2s.  2d.  2fqrs.    Ans. 


NOTE.  —  In  this  solution,  having  stated  the  question 
properly,  inverted  the  two  fractions  in  the  first  term  and 
then  arranged  all  the  terms  to  be  multiplied  into  each 
other,  I  take  the  shorter  method  of  cancelling  equal  nu- 
merators and  denominators,  and  abridging  such  as  are  di- 
visible by  the  same  number.  The  12  cancels  the  2  arid 
the  6,  and  the  5  divides  the  15,  leaving  a  denominator  3 
to  be  multiplied  into  the  only  remaining  denominator, 
which  is  3,  and  whose  product  is  9  ;  then  the  only  nu- 
merator above  unity  which  is  left,  is  55  ;  and,  consequent- 
ly, the  whole  is  brought  to  the  single  improper  fraction  -%5-, 
which  is  the  answer  required.  And  thus  let  every  simi- 
lar question  be  managed. 

2.  In  what  time  will  $350  gain  $26^  at  5  per  cent. 
per  annum  ?  Ans.  18  months. 

NOTE.-—  For  Double  Rule  of  Three,  Fellowship,  &e. 
in  decimals,  let  the  scholar  take,  as  exercises,  any  such 
questions  in  these  rules,  in  whole  numbers,  as  have  frac- 
tions or  compound  numbers  in  them. 

SIMPLE  INTEREST  BY  DECIMALS. 
A  TABLE  OF  RATIOS. 


Rate 

per  cent. 

Ratio.  |  Rate  per  cent. 

Ratio. 

3 

,03 

4* 

,055 

4 

,04 

6 

,06 

*i 

,045 

6j- 

,065 

5 

,05 

7 

,07 

Ratio  is  the  simple  interest  of  £1  for  1  year,  or,  in 
Federal  Money,  of  $1  for  one  year,  at  the  rate  percent, 
agreed  on. 

RULE. — Multiply  the  Principal,  Ratio,  and  Time  con- 
tinually together,  and  the  last  product  will  be  the  interest 
required* 


164  SIMPLE    INTEREST   BY   DECIMALS. 

EXAMPLES. 

I.  Required  the  interest  of  $425,65cts.  for  6  years,  at 
6  per  cent,  per  annum. 

$.  cts. 

425,65  Principal. 
,06  Ratio. 


25,53   90  Interest  for  one  year. 
6  Multiply  by  the  time. 


153,23  40  Ans.«=$153,23cts.  4m. 

2.  What  is  the  interest  of  .£645  10s.  for  3  years,  at  6 
per  cent,  per  annum  ? 

£645,5x,06x3=<£l  16,190=^116  3s.  9d..2,4qrs.  Ans. 

3.  Required  the  amount  of  $648,50cts.  for  12J  years, 
at  5=5-  per  cent,  per  annum.  Ans.  $1103,26cts.+ 

4.  What  is  the  amount  of  $243,39cts.  for  1^  year  at  6 
per  cent,  per  annum  ?  Ans.  8270,7451. 

CASE  2. —  The    amount,  time,  and    ratio    given,  to  Jind 
the  principal. 

RULE. — Multiply  the  ratio  hy  the  time  ;  add  unity  to 
the  product  for  a  divisor,  by  which  sum  divide  the  amount, 
and  the  quotient  will  be  the  principal. 

EXAMPLES. 

1.  What  principal  will  amount  to  $1235,97cts.  5m.  in 

5  years,  at  6  per  cent,  per  annum  1 

__  <s>  <& 

,06  x 5 + 1 = 1 ,30)  1235,975(950,75  Ans. 

2.  What  principal  will  amount  to  £956   10s.  4,125d. 
in  8f  years,  at  5^  per  cent.  1  Ans.  ,£645  15s. 

3.  What  principal  will  amount  to   $l3S4,50cts.  in  7 
years,  at  6  per  cent,  per  annum  ?  Ans.  $975. 

CASE    3. —  The   amount,   principal,   and  time    given,    to 
Jind  the  ratio. 

RULE. — Subtract  the  principal  from  the  amount;  divide 
the  remainder  by  the  product  of  the  time  and  principal, 
and  the  quotient  will  be  the  ratio. 


SIMPLE    INTEREST    BY   DECIMALS.  165 

EXAMPLES. 

1.  At  what  rate  per  cent,  will  $950,75   amount  to 
$1235,975  in  5  years  ? 

From  the  amount =1235,975 
Take  the  principal =950,75 

950,75x5=4753,75)285,2250(,06=6  per  cent.  Ans. 

2.  At  what  rate  per  cent,  will  £543  amount  to  £705 
18s.  in  5  years  ?  Ans.  6  per  cent. 

3.  At  what  rate  per  cent,  will   $2124,25  amount  to 
$3482,44234375  in  7f  years  ?  Ans.  8£  per  cent. 

CASE  4. —  The   amount,   principal,    and    rate   per    cent. 

given, .to  Jind  the  time. 

RULE.-— Subtract  the  principal  from  the  amount ;  di- 
vide the  remainder  by  the  product  of  the  ratio  and  prin- 
cipal ;  and  the  quotient  will  be  the  time. 

EXAMPLES. 

1.  In  what  time  will  $248,39  amount  to  $270,7451  at 
6  per  cent,  per  annum  1 

From  the  amount     $270,7451 
Take  the  principal     248,39 

248,39x,06= 14,9034)22,355 1(1, 5  =  !£  year.    Ans. 

2.  In  what  time  will  c£543  amount  to  "^705  18s.   at  6 
per  cent,  per  annum  ?  Ans.  5  years. 

'3.  In  what  time  will  $2142,25  amount  to  $3482,4423- 
4375  at  8J  per  cent,  per  annum  1  Ans.  7f  years. 

TO  CALCULATE  INTEREST  FOR  DAYS. 

A  TABLE  OF  RATIOS  FOR  DAYS. 


Rate  per  cent.        R.atios. 
4^=,0001232S767 
5~=, 000 1 369863 
5£=,00015068493 


Rate  per  cent.  Ratios. 
6  =,00016438356 
6i=,00017808219 
7"  =,00019178082 


Rule. — Multiply  the  principal  by  the  given  number  of 
days  and  that  product  by  the  ratio  for  a  year  ;  divide  the 
last  product  by  365,  (the  number  of  days  in  a  year,)  and  it 


166  SIMPLE  INTEREST  BY  DECIMALS. 

will  give  the  interest  required.  Or,  multiply  the  ratio 
for  a  day  in  the  foregoing  table  by  the  principal,  and  that 
product  by  the  given  number  of  days  ;  and  the  last  pro- 
duct will  be  the  interest  required. 

EXAMPLES. 

1.  What  is  the  interest  of  ^300,  10s.  for  146  days,  at 
6  per  cent.  ? 

360,5xl46x,06    £.          £.  s.  d.  qr. 
---  =8,652=8  13  0  1,92  Ans. 

365 
Or,  00016438356x360,5x146  ==£8,6519999+Ans. 

2.  What  is  the  interest  of  g780,40cts.  for  100  days  at 
6  per  cent,  per  annum  ?  Ans.  $12,S2cts.  8m.  + 

3.  What  is  the  interest  of  S48l,75cts.  for  25  days  at  7 
percent,  per  annum  ?  Ans.  $2,30cts.  9m.  + 

NOTE.  —  The  interest  of  any  sum  for  6  days,  at  6  per 
cent,  is  just  as  many  mills  and  decimals  of  a  mill,  as  the 
principal  contains  dollars  and  decimals  of  a  dollar. 
Therefore  set  down  the  principal,  multiply  it  by  the  days, 
and  divide  the  product  by  6  ;  the  quotient  will  be  the  in- 
terest in  mills  and  decimals  of  a  mill.  This  is  calling 
only  30  days  a  month. 

What  is'the  interest  of  g231,84  for  100  days  ? 
231,  84x1  00  days? 


6  Or  $3,86  4  ;  which 

is  5cts.  1m.  too  much  ;  but  when  the  time  is  less  than  30 
days,  it  gives  the  answer  very  exact,  for  ordinary  sums. 

When  interest  is  to  be  calculated  on  cash  account-:, 
&c.  where  partial  payments  are  made,  it  is  the  common 
practice  to  multiply  the  several  balances  into  the  days 
they  are  at  interest  ;  then  to  multiply  the  sum  of  these 
products  by  the  rate  on  the  dollar,  and  divide  the  last 
product  by  365  ;  and  thus  cast  the  whole  interest  due  on 
the  account,  &c. 

EXAMPLE. 

Lent  John  Joy,  per  bill  on  demand,  dated  1st  of  June, 
1821,  $2000,  of  which  I  received  back  the  19th  of  August, 
$400  ;  on  the  15th  of  October,  $600;  on  the  llth  of  De- 
cember, §400  ;  on  the  l7th  of  February,  1822,  $200  ; 
and  on  the  1st  of  June,  $400  ;  how  much  interest  is  due 
on  the  bill,  reckoning  at  6  per  cent.  ? 


COMPOUND    INTEREST   BY   DECIMALS. 

1821. 

June  1.  Principal,  per  bill, 
Aug.  19.  Received  in  part, 

Balance, 
Oct.  15.  Received  in  part 

Balance, 
Dec.  11.  Received  in  part, 

1822,  Balance, 

Feb.  17.  Received  in  part, 

Balance, 

June  1.  Rec'd  in  full  of  principal,   400 
Then  388600 

,06  Ratio. 

$  cts.m. 

365)23316,00(63,879+  Ans.  =63,87  9 


167 


$ 

2000 
400 

days. 
79 

57 

products. 
158000 

91200 

1600 
600 

1000 
400 

57 

57000 

600 
200 

68 

40800 

400 
,  400 

104 

41600 

388600 

COMPOUND  INTEREST  BY  DECIMALS. 


A  table  showing  the  amount  of  £1  or  $1  at  5  and  6  per 

cent,  per  annum,  compound  interest,  for  20  years. 

Yrs. 

5  per  cent. 

6  per  ct. 

Yrs. 

5  per   cent. 

6  per    cent. 

1 

,05000 

1,06000 

11 

1,71033 

1,89829 

2 

,15250 

1,12360 

12 

1,79585 

2,01219 

3 

,15762 

1,19101 

13 

1,88564 

2,13292 

4 

,21550 

1,26247 

14 

1,97993 

2,26090 

5 

,27628 

1,33822 

15 

2,07892 

2,39^55 

6 

,34009 

1,41851 

16 

2,18287 

2,54035 

7 

,40710 

1,50363 

17 

2,29201 

2,69277 

8 

,47745 

1,59384 

18 

2,40661 

2,85433 

9 

,55132 

1,68947 

19 

2,52695 

3,02559 

10 

1,62889 

1,79084  j  20 

2,65329 

3,20713 

RULE. — Multiply  the  given  principal  continually  by  the 
amount  of  one  pound,  or  one  dollar,  for  one  year,  at  the 
rate  per  cent,  given,  until  the  number  of  multiplications 
is  equal  to  the  given  number  of  years,  and  the  product 
will  be  the  amount  required. 


168  ANNUITIES    AT    COMPOUND    INTEREST. 

Or,  Take  from  the  preceding  Table,  the  amount  of  one 
pound,  or  one  dollar,  as  the  case  may  be,  for  the  given 
number  of  years,  and  at  the  given  rate  per  cent.,  and 
multiply  it  by  the  given  principal,  and  it  will  give  the 
amount  as  before. 

EXAMPLES. 

1.  What  will  £400  amount  to  in  4  years,  at  6  per  cent, 
per  annum,  compound  interest? 

400xl,06xl,06xl,06xl,06=£504,99+  or  ,£504  19s. 

9d.  2,4qrs.-f  Ans. 
Or,  by  the  Table.     Tabular  amount  of  ,£1  =  1,26247 

Multiply  by  the  principal  400 

Whole  amount  £504,98800 

2.  What  is  the  compound  interest  of  §555  for  14  years, 
at  5  per  cent.  ?  Ans.  §543,S6cts.+ 

NOTE. — Any  sum  of  money,  at  6  per  cent,  per  annum, 
simple  interest,  will  double  in  16f  years;  but  at  6  per 
cent,  per  annum  compound  interest,  it  will  double  in  11 
years  and  325  days,  or  11,889  years. 

ANNUITIES  AT  COMPOUND  INTEREST. 

CASE  1. —  To  find  the,  amount  of  an  annuity,   fyc. 

RULE. — Raise  the  amount  of  $1,  or  JBl,  at  the  given 
rate  per  cent.,  for  one  year,  to  that  poicer  denoted  by  the 
given  number  of  years  ;  subtract  unity  or  1  from  this  pro- 
duct ;  multiply  the  remainder  by  the  given  annuity  ;  di- 
vide this  last  product  by  the  ratio  made  less  by  unity  or 
1  ;  and  the  quotient  will  be  the  amount  sought. 

EXAMPLES. 

1.  If  $250,  yearly  pension,  be  foreborne  7  years,  what 
will  it  amount  to,  at  6  per  cent,  per  annum  compound  in- 
terest ? 


1,06x1,06x1,06x1,06x1,06x1,06x1,06— I,x250 

1,06—1, 

$2098,45c.  9m. -f  Ans. 


INVOLUTION.  169 

2.  If  a  salary,  or  an  annuity,  .of  ^100  per  annum,  runs 
on  unpaid  for  6  years,  at  5  per  cent,  compound  interest, 
what  is  the  amount  due  at  the  end  of  that  period  ? 

Ans.  «£(iSO  3s.  9£d.  ,63. 

CASE  2.  —  To  find  the  present  worth   of  an  Annuity,   fye. 
RULE.  —  Raise  the  amount  of  gl,  or  £l,  at  the  given 
rate  per  cent.,  for  1  year,  to  that  power  denoted  by  the 
given  number  of  years  ;  divide  the  given  annuity  by  this 
product;  subtract  its  quotient  from  the   given  annuity; 
divide  the  remainder  by  the  ratio  made  less  by  unity  or 
1  ;  and  the  quotient  will  be  the  present  worth  sought. 
EXAMPLES. 

1.  What  is  the  present  worth  of  a  salary  of  $300,  to 
continue  5  years,  at  5  per  cent,  compound  interest  ? 

300 

----  =235,0578499405.  + 
1,05x1,05x1,05x1,05x1,05 

300—235,0578499405         g     c.  m. 
Then,  ---  =1298,84  3+Ans. 
1,05—1, 

2.  What  is  the  present  worth  of  ^630  per  annum,  to 
continue  7  years,  at  6  per  cent,  compound  interest  1 

Ans.  ^'167  9s.  5d.+ 


INVOLUTION. 

INVOLUTION  is  the  continual  multiplication  of  a  num- 
ber into  itself;  and  the  products  thence  arising,  with  the 
original  number  itself,  are  called  the  powers  of  that  num- 
ber. 

Any  number  may  itself  be  called  a  first  power.  If  the 
first  power  be  multiplied  by  itself,  the  product  is  called 
the  second  power,  or  square  ;  if  the  square  be  multiplied 
by  the  first  power,  the  product  is  called  the  third  power, 
or  cube  ;  if  the  cube  be  multiplied  by  the  first  power,  the 
product  is  called  the  fourth  power,  or  biquadrate,  &,e. 

Thus  3  is  the  first  power  of  3. 
3x3=9  is  the  second  power  of  3. 
3x3x3=27  is  the  third  power  of  3. 
3x3x3x3=81  is  the  fourth  power  of  3  &c.  <fcc. 
And  in  this  manner  is  formed  the  folio  wing  table  of  powers. 


170    EVOLUTION,  OR  EXTRACTION  OF  ROOTS. 

Table  of  the  SQ  UARES  and  CUBES  of  tlie  nine  digits. 


Roots. 
Squares. 
Cubes. 

1 

2 
4 

8 

3 
9 

27 

4 
16 
64 

5 

6 

7 

8 

9 

1 
1 

25 

36 

49 

64 

81 

125 

216 

343 

512 

729 

EXAMPLES. 

1.  What  is  the  6th  power  of  8? 
8     the  root,  or  1st  power. 

8 


64  = 

8 


=2d  power,  or  square. 


NOTE. — The  num- 
ber denoting  the 
height  of  the  power 
is  called  the  index, 
or  exponent  of  that 
power;  so  the  2d 
power  of  3  may  be 
denoted  by  32,  the  3d 
by  33,  the  4th  by  34, 

&c. ;  the  6th  power 

4096 =4th  power, or  biquadrate.  of  8  by86,  &c. 
8 


5l2=3d  power,  or  cube. 

8 


32768  =5th  power,  or  sursolid. 
8 

262144=6th  power,  or  square  cube.  Ans.  86. 

2.  What  is  the  7th  power  of  f  ?  Ans. 

3.  What  is  the  5th  power  of  |  or  l£  ?     Ans. 

4.  What  is  the  fourth  power  of  ,27  ?     Ans.  ,00531441. 


EVOLUTION,  OR  EXTRACTION  OF  ROOTS. 

WHEN  the  root  of  any  power  is  required,  the  business 
of  finding  it  is  called  the  extraction  of  the  Root. 

The  root  is  that  number,  which  by  a  continual  multi- 
plication into  itself,  produces  the  power  which  is  given  to 
be  extracted. 

%Though  every  number  will  produce  a  perfect  power  by 
involution,  yet  there  are  many  numbers,  the  precise  roots 
of  which  can  never  be  determined.  By  the  help  of  deci- 
mals, however,  we  can  approximate  towards  the  root,  to 
any  assigned  degree  of  exactness. 


EXTRACTION  OF  THE  SQUARE  ROOT.       171 

The  roots  which  approximate,  are  called  surd  roots, 
and  those  which  are  perfectly  accurate,  are  called  rational 
roots, 

TO  EXTRACT  THE  SQUARE  ROOT. 

Any  number  multiplied  into  itself,  produces  a  square. 
The  extracting  of  the  square  root,  is  only  finding  a  num- 
ber, which,  being  multiplied  into  itself,  shall  produce  the 
given  number. 

RULE. — 1.  Distinguish  the  given  number  into  periods 
of  two  figures  each,  by  putting  a  point  over  the  place  of 
units,  another  over  the  place  of  hundreds,  and  so  on  over 
every  second  figure  ;  and  if  there  be  decimals,  point  them 
in  the  same  manner,  from  units  towards  the  right  hand  ; 
which  points  show  the  number  of  figures  the  root  will  con- 
sist of. 

2.  Find  by  the  table  or  trial  the  greatest  square   num- 
ber in  the  first,  or  left  hand  period,  place  the  root  of  it  at 
the  right  hand  of  the  given  number,  (after  the  manner  of 
a  quotient  in  division,)  for  the  first  figure  of  the  root ; 
and  set  the  square  number  under  the  period,  subtract  it 
therefrom,  and  to  the  remainder  bring  down  the  next  pe- 
riod for  a  dividend. 

3.  Place  the  double  of  the  root  already  found  on  the 
left  hand  of  the  dividend  for  a  divisor. 

4.  Consider  what  figure  must  be  annexed  to  the  divis- 
or, so  that  if  the  result  be  multiplied  by  it  the  product 
may  be  equal  to,  or  the  next  less  than,  the  dividend,  and 
it  will  be  the  second  figure  of  the  root. 

5.  Subtract  that  product  from  the  dividend,  and  to  the 
remainder  bring  down  the  next  period  for  a  new  dividend. 

6.  Find  a  divisor  as  before,  by  doubling  the  figures  al- 
ready in  the  root ;  and  from  these  find  the  next  figure  of 
the  root  as  in  the  last  article  ;  and  so  on  through  all  the 
periods  to  the  last. 

Or,  to  facilitate  the  foregoing  operation,  when  a  period 
is  brought  down  to  a  remainder,  and  a  dividend  thus  form- 
ed, in  order  to  find  a  new  figure  in  the  root,  divide  said 
dividend,  (omitting  the  right  hand  figure  thereof,)  by  do*u- 
ble  the  root  already  found,  and  the  quotient  will  common- 
ly  be  the  figure  of  the  root  sought,  or,  being  made  less  by 
one,  or  two,  will  generally  give  the  next  figure  sought, 


172      EXTRACTION  OF  THE  SQUARE  ROOT. 

To  extract  the  square  root  of  a  Vulgar  Fraction. 
First  prepare  all  vulgar  fractions  by  reducing  them  to 
their  lowest  terms,  both  for  this  and  all  other  roots.  Then, 

1.  Take  the  root  of  the  numerator  and  that  of  the  de- 
nominator for  the  respective  terms  of  the  root  required  ; 
and  this  is  the  best  way  if  the  denominator  be  a  complete 
power.     But  if  not, 

2.  Multiply  the  numerator  and  denominator  together  ; 
take  the  root  of  the  product  ;  this  root,  being  made  the 
numerator  to  the  denominator  of  the   given   fraction,   or 
the  denominator  to  the  numerator  of  it,  will  form  the 
fractional  root  required. 

3.  Or  raduce  the  vulgar  fraction  to  a  decimal,  and  ex- 

tract its  root. 

EXAMPLES. 
1.  Required  the  square  root  of  6749604. 

6749604(2598  Ans.     The  root  exactly  without  a 
4  remainder  ;  but  when  the  periods  be- 

-  -  longing  to  any  given  number  are   all 

45)274  exhausted,  arid  still  leave  a   remain- 

5)225  der,  the  operation  may  be  continued 

---  at  pleasure,  by  annexing  periods  of 

509)4996  ciphers,  &c.  —  Roots  are  often  denot- 

9)4-581  ed  by  writing  v/   before  the  power, 

with  the  index  of  the  root  within   or 
5188)41504  over  it,  save  the  index  of  the   square 

8)41504  root,  which  is  ever  understood  :  so  v/ 

64  is  the  2d  or  square  root  of  64  ; 


2,  Required  the  64  the  3d  or  cube  root  of  64  ;>  </  64 
square  root  of  739,4  the  4th  root  of  64. 

739,40(27,19  +  root.  . 

4  When  the  square  root  of  a  number 

is  wanted  to  many  places,  the  work 
47)339  may  be  much   abridged.      Find  half 

7)329  the  root  by  the  rule  ;  then  to  get  the 

rest,  annex  to  the  last  remainder  as 


541)   1040  many  ciphers  as  you  need,  and  divide 

1)     541  it  by  the  double  of  the  root   before 

found. 


5429)49900 
9)48861 


1039  remainder. 


APPLICATION  AND  USE  OF  THE  SQUARE  ROOT.          173 

3.  What  is  the  square  root  of  2  ?  Ans.  1,41421356.+ 

4.  What  is  the  square  root  of  10342056?  Ans.  3216. 

5.  What  is  v/964,5192360241  ? 

Ans.  31,05671. 

6.  What  is  v/,00032754? 

Ans.  ,01809.+ 

7.  What  is  \/fHt-  Ans-  rV 

8.  What  is  x/42J  ?  Ans.  8f 

9.  What  is  x/6f  1     Ans.2,5298+&c. 


APPLICATION  AND  USE  OF  THE  SQUARE 
ROOT. 

CASE  1.  —  To  Jind  a  mean  proportional  between  any  two 

given  numbers. 

RULE.  —  Multiply  the  two  given  numbers  together,  and 
extract  the  square  root  of  the  product,  which  root  will 
be  the  number  sought. 

EXAMPLE. 
What  is  the  mean  proportional  between  16  and  36  ! 

v/36x!6=24.  Ans. 

CASE  2.  —  To  find  the  side  of  a  square  equal  in    area  to 
any  given  superficies. 

RULE.  —  Extract  the  square  root  of  the  given  superficies* 
which  root  will  be  the  side  of  the  square  sought. 

EXAMPLES. 

1.  If  an  acre  of  land  contains  160  square  rods,  what 
will  be  the  side  of  a  square,  which  should  contain  just  aa 
acre  ?  v/  160  =  12,649  +  rods.  Ans. 

2.  A  general  having  an  army  of  5184   men   wishes  to 
form  them  into  a  square  ;  how  many  must   he  place   ia 
rank  and  file  !  v/5  1  84  =72.  Ans. 

3.  Let  8192  men  be  formed  into  an  oblong,  so  that  the 
mmvber  in  rank  may  be  double  the  file. 

8192 

-v/  -  =64  in  fite.  64x2=123  io  rank. 
2 
P2 


174   APPLICATION  AND  USE  OF  THE  SQUARE  ROOT. 

4.  Suppose  a  gentleman  would  set  out  an  orchard  of 
864  trees,  so  that  the  length  shall  be  to  the  breadth  as  3 
to  2,  and  the  distance  of  each  tree,  one  from  the  other,  7 
yards  ;  how  many  trees  must  there  be  in  length,  and  how 
many  in  breadth,  and  how  many  square  yards  of  ground 
do  they  stand  on  1 

To  resolve  any  question  of  this  nature,  say,  as  the  ra- 
tio in  length  is  to  the  ratio  in  breadth,  so  is  the  number  of 
trees  to  a  fourth  number,  whose  square  root  is  the  number 
in  breadth  ;  then  as  the  ratio  in  breadth  is  to  the  ratio  in 
length,  so  is  the  number  of  trees  to  a  fourth  number, 
whose  root  is  the  number  in  length.  And  as  unity  is  to 
the  distance,  so  is  the  number  in  length  less  by  one  to  a 
fourth  number ;  next  do  the  same  by  the  breadth,  and 
multiply  the  two  numbers  thus  found  together,  and  the 
product  will  be  the  answer. 

As  3  :  2  :  :  864  :  576,&  ^7576=24  num.  in  breadth.  Ans. 
As  2  :  3  :  :  864  :  1 296, &V 1 296 =36  num.  in  length  Ans. 
As  1  :  7 : :  36—1  :  245,     And,  as  1  :  7  :  :  24—1  :  101. 
And  245x161=39445  square  yards.  Ans. 

CASE  3. —  To   ascertain   the  proportionate   capacities    of 

water  pipes. 

RULE. — Square  the  given  diameter,  and  multiply  said 
square  by  the  given  proportion  ;  the  square  root  of  the 
product  is  the  answer. 

EXAMPLE. 

Admit  lOhhds.  of  water  are  discharged  through  a  leaden 
pipe  of  2J-  inches  diameter,  in  a  certain  time  ;  what  must 
be  the  diameter  of  another  pipe,  that  shall  discharge  four 
times  as  much  water  in  the  same  time  ? 

2!=2,5  and  2,&x 2,5  =« 6,25  square. 

4  given  proportion. 


v/25,00=5  inches  diam.  Ans. 

CASE  4.—  The  sum  of  any  two  numbers,  and  their  product 

being  given ,  to  find  each  number. 

RULE. — From  the  square  of  their  sum,  subtract  4  times 
their  product,  and  extract  the  square  root  of  the  remain- 


APPLICATION  AND  USE  07  THE  SQUARE  ROOT.  175 

der,  which  will  be  the  difference  of  the  two  numbers  ; 
then  half  the  said  difference  added  to  half  the  sum,  gives 
the  greater  of  the  two  numbers,  and  the  said  half  differ- 
ence, subtracted  from  the  half  sum,  gives  the  less  number. 

EXAMPLES. 

1.  The  sum  of  two  numbers  is  46,  and  their  product 
is  504  ;  what  are  those  two  numbers  1 

The  sum  of  the  numbers  46x46=2116  sq.  of  their  sum. 
The  product  of  ditto.  504x4=2016  four  times  the  pro. 

[numbers. 

^/100  =  10         differ,    of   the 

46—2=23  half  sum.  10-f-2=5  half  differ. 

+5  half  diff.  23  half  sum. 

—5  half  differ. 
28  greater  number.    Ans. — 

18  less  number.  Ans. 

2.  Bought  a  certain  quantity  of  broadcloth  for   $573, 
75cts. ;  and  if  the  number  of  cents  which  it  cost  per 
yard,  was  added  to  the  number  of  yards  bought,  the  sum 
would  be  480  ;  how  many  did  I  buy,  and  at  what  price 
per  yard.  Ans.  255yds.  at  $2,25cts.  per  yard. 

3.  If  I  lay  out  a  lot  of  land  in   an  oblong  form,  con- 
taining 7  acres,  1  rood,  and  10  rods,  and  taking  just  142 
rods  of  wall  to  enclose  it  ;  pray  how  many  rods  long,  and 
how  many  wide  is  said  lot  ? 

Ans.  45  rods  long,  and  26  rods  wide. 

CASE  5. —  To  Jind  the  degree  of  light,  heat,  or  attraction. 

NOTE. — The  effects  on  degrees  of  light,  heat,  and  attrac- 
tion, are  in  proportion*  to  the  squares  of  the  distance, 
whence  they  are  propagated. 

EXAMPLES. 

1.  Two  men,  A  and  B,  are  sitting  in  a  room,  the  for- 
mer 3,  and  the   latter   6  feet  distant  from  a  fire ;   how 
much  hotter  is  it  at  A's,  than  at  B's  seat  ? 
3x3=9,&  6x6=36.      Then,  as  9 :  1  :  :  36  :  4r  so  that 

A's  place  is  4  times  as  hot  as  B's.     Ans. 

2.  If  the  earth's  mean  distance  from  the  sun  be  95,000- 
000  of  miles,  at  what  distance  from  him  must  another 


176   APPLICATION  AND  USE  OF  THE  SQUARE  ROOT. 

body  be  placed,  that  it  may  receive  a  degree  of  light  and 
heat,  double  to  that  of  the  earth  1 

95  OOOOOO2 

\/ =67175144  + mile.  Ans.  which  is  somewhat 

2         less  than  the  distance  of  Venus  from  the  sun. 

3.  A  ball  descending  by  the  force  of  gravity  from  the 
top  of  a  tower,  was  observed  to  fall  half  the  way  in  the 
last  second  of  time ;  how  long  was  it  in  descending,  and 
what  was  the  height  of  the  tower  ? 

The  square  roots  of  the  distance  are  as  the  times,  viz. 
As  the  5/1  :  -v/2  :  :  the  time  of  falling  through  :  the  whole 
required  height. 

Now,  the  x/l  =  l,  and  ^2  =  1,4142,  from  which  take 
1 ;  ,4142  remains. 

And,  as  ,4142  :  1,4142  :  :  1  :  3,414  +  sec.  time  of  de- 
scent ;  the  square  of  which  is  11,6554  nearly.  And  the 
velocity  acquired  by  heavy  bodies  falling  near  the  surface 
of  the  earth,  is  16  feet  in  the  first  second,  64  in  the  second 
second,  144  in  the  third  second,  &/c.  that  is,  the  space 
fallen  through  [in  feet]  is  always  equal  to  the  square  of 
the  time  in  4ths  of  a  second, 
ft.  sec.  sq. 

As  I2  :  16  :  :  11,6554  :  186,4864  =  186^  feet  nearly, 
height  of  the  tower,  Ans. 

CASE  6. — Any  two  sides  of  a  right-angle  triangle  given 
to  find  the  other  side. 

RULE. — Extract  the  square  root  of  the  sum  of  the 
squares  of  the  two  least  sides,  and  that  root  is  the  great- 
est side  ;  for  the  square  root  of  the  sum  of  the  squares  of 
the  two  legs,  is  always  the  length  of  the  hypotenuse. 
Extract  the  square  root  of  the  difference  of  the  squares 
of  either  of  the  two  least  sides  and  the  greatest  side,  and 
that  root  is  the  other  side  ;  for  the  square  root  of  the  dif- 
ference of  the  squares  of  either  leg  and  the  hypotenuse,  is 
always  the  length  of  the  other  leg. 

EXAMPLES. 

1.  A  ladder  40  feet  long  may  be  so  planted  as  to  reach 
a  window  33  feet  from  the  ground,  on  one  side  of  the 
street;  and,  without  moving  it  at  the  foot,  will  do  the 
same  by  a  window  21  feet  high  on  the  other  side;  how 
wide  is  the  street  ? 

402=xl600.  332=1089.  212=441.  Then  1600—1089 


EXTRACTION  OF  THE  CUBE  ROOT.       177 

=511,  and  x/51  1=22,6;    and  1600—441=1159,  and 
v/1  159=34,04;  then  22,6+34,04=54,64  feet.+     Ans. 

2.  A  line  27  yards  long  will  exactly  reach   from  the 
top  of  a  fort  to  the  opposite  bank  of  a  river,  known  to  be 
23  yards  broad  ;  what  is  the  height  of  the  wall  ? 

Ans.  14,142+yards. 

3.  Two  ships  sail  from  the  same  port  ;  one   sails   due 
east  50  leagues,  and  the  other  due  north  84  leagues  ;  how 
far  are  they  then  apart  ?  Ans.  97,75+  leagues. 


TO  EXTRACT  THE  CUBE  ROOT. 

A  cube  is  any  number  multiplied  by  its  square. 

To  extract  the  Cube  Root,  is  to  find  a  number  which, 
being  multiplied  into  its  square,  shall  produce  the  giveji 
number. 

RULE.  —  1.  Separate  the  given  number  into  periods  of 
three  figures  each,  by  putting  a  point  over  the  unit  figure, 
and  every  third  figure  both  ways  from  the  place  of  units. 

2.  Find  the  nearest  less  cube  to  the  first  period  by  the 
table  of  powers  or  trial  ;  set  its  root  in  the  quotient  ;  sub- 
tract the  cube  found  from  the  first  period,  and  to  the  re- 
mainder bring  down  the  second  period,  and  call  this  the 
resolvend. 

3.  To  three  times  the  square    of  the  root  just  found, 
add  three  times  the  root  itself,  setting  this  one  place  more 
to  the  right  than  the  former,  and  call  this  sum  the  divisor. 
Then  divide  the  resolvend,  omitting  the  unit  figure,  by 
the  divisor,  for  the  next  figure  of  the  root,  which    annex 
to  the  former,  calling  this  last  figure   c,  and  the  part  of 
the  root  before  found  call  a. 

4.  Add  together  these  three  products,  viz.  thrice  the 
square  of  a  multiplied  by  e,  thrice  a  multiplied    by  the 
square  of  e,  and  the  cube  of  e,  setting  each  of  them  one 
place  more  to  the  right  hand  than  the   former,   and  call 
the  sum  the  subtrahend,  which  must  not  exceed  the  resol- 
vend ;  but  if  it  do,  then  make  the  last  figure  e   less,  and 
repeat  the  operation  for  finding  the  subtrahend. 

5.  From  the  resolvend  take  the  subtrahend,  and  to  the  • 
remainder  join  the  next  period  of  the  given  number  for  a 


178   APPLICATION  AND  USE  OF  THE  CUBE  ROOT. 

new  resolvend  ;  to  which  form  a  new  divisor  from  the 
whole  root  now  found,  and  thence  another  figure  of  the 
root  as  before,  &,c. 

EXAMPLES. 
1.  Required  the  cube  root  of  48228,544. 


3x32=27 
3x3  =  09 


Divisor     279 


48228,544(36,4  root. 

27 


21228  resolvend. 


3x32x6=162        ) 
3x3x62  =  324      VAdd. 
6*=     216  j 


3x362=3888 
3x36  =     108 

- 

19656  subtrahend. 

Divisor  38988 

1572544  resolvend. 

3x362x4  =         15552        V 
3x36  x42«             1728       }  Add. 

43=5                        64      J 

1572544  subtrahend. 

2.  What  is  the  cube  root  of  1092727?         Ans.  103. 

3.  What  is  ^34965783  ?     Ans.  327. 

4.  What  is  ^/,0001357  ? 

Ans.  ,05138.+ 

5.  What  is  -NXfriny  Ans-  §• 

6.  What  is  ^/f  ?  Ans.  ,873.+ 

7.  What  is  x3/-2?  Ans.  1,25.+ 

8.  What  is  •3'*i*  ?  Ans.  f. 


APPLICATION  AND  USE  OF  THE  CUBE  ROOT. 

EXAMPLES. 

1.  The  statute  bushel  contains  2150,4197+  cubic  or 
solid  inches  ;  I  demand  the  side  of  a  cubic  box  which 
shall  contain  just  that  quantity  ? 

^/2150,419724=12,907+inch.  Ans. 

NOTE. — The  solid  contents  of  similar  figures,  are  in 
proportion  to  each  other,  as  the  cubes  of  their  similar 
sides  or  diameters. 


TO  EXTRACT  THE  ROOTS  OF  POWERS  IN  GENERAL.  179 

2.  If  a  bullet  4  inches  diameter  weigh  9^.,  what  will  a 
bullet  of  the  same  metal  weigh,whose  diameter  is  8  inches? 
4x4x4=64  Sx8x8=»5l2.  As  64  :  9  :  :  512  :  72ft.  Ans. 

3.  If  a  solid  globe  of  silver,  of  3  inches  diameter,  be 
worth  $150,  what  is  the  value  of  another  globe  of  silver, 
whose  diameter  is  eight  inches  ? 

3x3x3=27.  8x8x8=512.  As  27  :  8150  :  :  512  : 

$2844J-f  Ans. 

The  side  of  a  cube  being  given,  to  find  the  side  of  that 
cube  which  shall  be  double,  triple,  &c.  in  quantity  to  the 
given  cube. 

RULE. — Cube  the  given  side,  and  multiply  it  by  the 
given  proportion  between  the  given  and  required  cube, 
and  the  cube  root  of  the  product  will  be  the  side  sought. 

4.  If  a  cube  of  silver,  whose  side  is  2  inches,  be  worth 
820,  what  should  the  side  of  a  cube   of  like  silver  be, 
whose  value  would  be  8  times  as  much  ? 

2x2x2=8,  and  8x8=64.     ^64=4  inches.    Ans. 

5.  There  is  a  cubical  vessel  whose  side  is  4  feet ;  I  de- 
mand the  side  of  another  cubical  vessel,  which  shall  con- 
tain 4  times  as  much  1 

4x4x4=64.  &  64x4=256.     -^256 =6,349+ feet.  Ans. 

6.  A  cooper  having  a  cask  40  inches  long,  and  32  in- 
ches at  the  bung  diameter,  is  ordered   to  make  another 
cask  of  the  same  shape,  but  which  shall  hold  just  twice 
as  much  ;  what  will  be  the  bung  diameter,  and  length  of 
the  new  cask  1 

40x40x40x2=128000;    then  ^/ 128000 =50,3+  in- 
ches length.  Ans. 

32x32x32x2=65536;       <fc     ^65536=40,3  +  inches 

bung  diam.  Ans. 


TO  EXTRACT  THE  ROOTS  OF  POWERS  IN  GENERAL. 

RULE. — 1.  Prepare  the  given  number  for  extraction  by 
pointing  oft"  from  the  place  of  units  as  the  root  required 
directs  ;  4th  root  put  a  dot  over  every  4th  figure  &,c.  from 
the  place  of  units ;  5th  root,  over  every  5th,  &c.  from 
units'  place,  &c. 

2.  Find  the  first  figure  of  the  root  by  trial,  and  sub- 
tract its  power  from  the  given  number. 


180  TO  EXTRACT  THE  ROOTS  OF  POWERS  IN  GENERAL. 

3.  To  the  remainder  bring  down  the  first  figure  in  the 
next  period,  and  call  it  the  dividend. 

4.  Involve  the  root  to  the  next  inferiour  power  to  that 
which  is  given,  and  multiply  it  by  the  number  denoting 
the  given  power*  for  a  divisor. 

5.  Find  how  many  times  the  divisor  maybe  had  in  the 
dividend, and  the  quotient  will  be  another  figure  of  the  root. 

6.  Involve  the  whole  root  to  the  given  power,  and  sub- 
tract it  from  the  given  number  as  before. 

7.  Bring  down  the  first  figure  of  the  next  period  to  the 
remainder  for  a  new  dividend,  to  which  find  a  new  divi- 
sor, and  so  on,  until  the  whole  is  finished. 

EXAMPLES. 
1.  What  is  the  cube  root  of  53157376  ? 

53 157376(376  root. 
27=33 

32x3=27)26i  dividend. 
50653 =37  3 


372x3=4107)25043  second  dividend. 

53157376=3763 

2.  What  is  the  biquadrate  root  of  34827998976  ? 

Ans.  431,9.+ 

3.  What  is  the  sursolid  root  of  281950621875  ? 

Ans.  195. 

4.  What  is  the  square  cubed,  or  sixth  root  of  16196, 
005304479729  ?  Ans.  5,03. 

5.  Find  the  seventh  root  of  34487717407,30751. 

Ans.  32,0 i.+ 

NOTE. — The  roots  of  most  powers  may  be  found  by  the 
square  and  cube  roots  only ;  therefore,  when  any  even 
power  is  given,  the  better  way  will  be,~  especially  in  very 
high  powers,  to  extract  the  square  root  of  it,  which  re- 
duces it  to  half  the  given  power,  then  the  square  root  of 
that  power ;  and  so  on  till  it  comes  to  a  square  or  cube. 

For  example,  suppose  a  12th  power  be  given ;  the 
square  root  of  that  reduces  it  to  a  sixth  power ;  and  the 
square  root  of  a  sixth  power  to  a  cube. 


ARITHMETICAL    PROGRESSION  181 

6.  Extract  the  eighth  root  of  7213895789838336. 

Ans.  96. 

7.  What  is  the  biquadrate  root  of  5308416  ?  Ans.  48. 


ARITHMETICAL  PROGRESSION. 

ANY  rank  of  numbers  more  than  two,  increasing  by  a 
common  excess,  or  decreasing  by  a  common  difference,  is 
said  to  be  in  Arithmetical  Progression :  such  are  the 
numbers  1,  2,  3,  4,  &c.  7,  5,  3,  1  ;  and  ,8,  ,6,  ,4,  ,2. 
When  the  numbers  increase,  they  form  an  ascending  series ; 
but  when  they  decrease,  they  form  a  descending  series. 

The  numbers  which  form  the  series,  are  called  the 
terms  of  the  progression. 

Any  three  of  the  five  following  terms  being  given,  the 
other  two  may  readily  be  found. 

1st.  The  first  term,    ) 

3d.   The  last  term,   }  commonl7  -Ca»ed  the  extremes. 

3d.    The  number  of  terms. 
4th.  The  common  difference. 
5th.  The  sum  of  all  the  terms. 

PROBLEM  1. 

The  first  term,  the  last  term,  and  the  number  of  terms  be- 
ing given,  to  Jind  the  sum  of  all  the  terms. 
RULE. — Multiply  the  sum  of  the  extremes  by  the  num- 
ber of  terms,  and  half  the  product  will  be  the  answer. 

EXAMPLES. 

1.  The  first  term  of  an  arithmetical  progression  is  1, 
the  last  term  21,  the  number  of  terms  11  ;  required  the 
sum  of  the  series. 

21 
1 

22  

11         21  +  lxll 

Or  =121  the  Ans. 

2)242  2 

Ans.  121 
Q 


182  ARITHMETICAL    PROGRESSION. 

2.  How  many  strokes  does  a  Venice  clock  strike  in  the 
compass  of  a  day,  going  to  24  o'clock  ?          Ans.  300. 

3.  If  100  stones  be  placed  in  a  right  line,   a  yard  dis- 
tant from  each  other,  and  the  first  a  yard  from  a  basket ; 
what  distance  will  that  man  go  who  gathers  them  up  sing- 
ly, returning  with  them  one  by  one  to  the  basket  1 

Ans.  5  miles  and  1300  yards. 

4.  A  draper  sold  100  yards  of  cloth  at5cts.  for  the  first 
yard,  lOcts.  for  the  second,  15  for  the  third,  &c.,  increas- 
ing 5cts.  for  every  yard  ;  what  did  the  whole  amount  to, 
and  what  did  it  average  per  yard  ? 

Ans.  Amount  was  $252^,  and  the   average  price  was 

$2,52cts.  5m.  per  yard. 

PROBLEM  II. 

The  extremes  and  the  number  of  terms  being  given,   to  find 

the  common  difference. 

RULE. — Divide  the  difference  of  the  extremes  by  the 
number  of  terms  less  1,  and  the  quotient  will  be  the 
common  difference. 

EXAMPLES. 

1.  The  extremes  are  3   and    19,    and   the   number  of 
terms  is  9  ;  required  the  common  difference,  and  the  sum 
of  the  whole  series. 

9     19          19+3x9 

I       3  And =99  the  sura. 

2 

8)    16 

2  difference. 

2.  A  man  is  to  travel  from  Boston  to  a  certain  place  in 
12  days,  and  to  go  but  three  miles  the  first  day,  increasing 
every  day  by  an  equal  excess,  so  that  the  last  day's  jour- 
ney may  be  58  miles ;  required  the  daily  increase,  and 
the  distance  of  the  place  from  Boston. 

Ans.  Daily  increase  5,  distance  366  miles. 
2.  A  man  had   12   sons   whose   several    ages   differed 
alike ;  the  eldest  was  49,  the  youngest  5  years  old  ;  what 
was  the  common  difference  of  their  ages  ?  Ans.  4  years. 


GEOMETRICAL    PROGRESSION.  183 

PROBLEM  III. 
Given  the  first  term,  the  last  term,  and  the  common   differ- 

ence,  to  find  the  number  of  terms. 

RULE.  —  Divide  the  difference  of  the  extremes  by  the 
common  difference,  and  the  quotient  increased  by  1,  is 
the  number  of  terms  required. 

EXAMPLES. 

1.  The  extremes  are  3  and  19,  and  the  common  differ- 
ence 2  ;  what  is  the  number  of  terms  ? 

19 
3 

2)16       19—3 

—  Qr  --  f.  i  —9  the  Answer. 
8  2 

1 

Ans.     9 

2.  Suppose  a  man  travel  the  first  day  7  miles,  the  last 
51  miles,  and  increase  his  journey  each  day  by  4  miles  ; 
how  many  days  will  he  travel,  and  how  far  ? 

Ans.  12  days,  and  343  miles. 


GEOMETRICAL  PROGRESSION. 

ANY  series  of  numbers,  the  terms  of  which  gradually 
increase  or  decrease  by  a  constant  multiplication  or  di- 
vision, are  said  to  be  in  Geometrical  Progression.  Thus, 
4,  8,  16,  32,  64,  &c.  and  81,  27,  9,  3,  1,  &c.  are  series 
in  geometrical  progression,  the  one  increasing  by  a  con- 
stant multiplication  by  2,  and  the  other  decreasing  by  a 
constant  division  by  3. 

The  number  by  which  the  series  is  constantly  increas- 
ed or  diminished,  is  called  the  ratio. 

PROBLEM  I. 
Given  the  first  term,  the  last  term,  and  the   ratio,  to  find 

the  sum  of  the  series. 

RULE.  —  Multiply  the  last  term  by  the  ratio,  and  from 
the  product  subtract  the  first  term,  and  the  remainder  di- 
vided by  the  ratio  less  1,  will  give  the  sum  of  the  series. 


184  GEOMETRICAL    PROGRESSION. 

EXAMPLES. 

1.  The  extremes  of  a  geometrical  progression  are  1  and 
65536,  and  the  ratio  4  ;  what  is  the  sum  of  the  series  ? 
65536 
4 


262144        4x65536—1 

1  Or =87381  Ans. 

4—1 
4—1=3)262143 

87381  Ans. 

2.  A  man  travelled  6  days  ;  the  first  day  he   went  4 
miles,  and  doubling  his  travel  each  day,  his  last  day's  ride 
was  128  miles ;  how  far  did  he  go  in  the  whole  ? 

Ans.  252  miles. 

3.  The  extremes  of  a  geometrical  series  are  1024  and 
59049,  and  the  ratio  is  l£  ;  what  is  the  sum  of  the  series  1 

Ans.  175099. 

PROBLEM  II. 

Given  the  first  term  and  the  ratio,  to  find  any  other  term 
assigned.* 

CASE  1. — When  the  first  term  of  the  series,  and  the  ratio, 
are  equal.^ 

RULE. — 1.  Write  down  a  few  of  the  leading  terms  of 
the  series,  and  place  their  indices  over  them,  beginning 
the  indices  with  a  unit  or  1. 

2.  Add  together  such  indices  as,  in  their  sum,  shall 
make  up  the  entire  index  to  the  term  required. 

*  As  the  last  term  in  a  long1  series  of  numbers  is  very  tedious  to 
be  found  by  continual  multiplication,  it  will  be  necessary  for 
more  readily  finding-  it  out,  to  have  a  series  of  numbers  in  arith- 
metical proportion,  called  indices,  whose  common  difference  is  1. 

f  When  the  first  term  of  the  series,  and  the  ratio,  are  equal,  the 
indices  must  beg-in  with  a  unit,  and  in  this  case,  the  product  of 
any  two  terms  is  equal  to  that  term,  signified  by  the  sum  of  their 
indices. 

rp,          (  1,  2,  3,     4,     5,  &c.     Indices  arithmetical  series. 
:  \  2,  4,  8,  16,  32,  &c.     Geometrical  series. 

Now,  2+3=5,  the  index  of  the  fifth  term,  and  4x8= 
32=the  fifth  term. 


GEOMETRICAL    PROGRESSION.  183 

3.  Multiply  together  the  terms  of  the  geometrical  se- 
ries belonging  to  those  indices,  and  the  product  will  be 
the  term  required. 

EXAMPLES. 

1.  The  first  term  of  a  geometrical  series  is  2,  and  the 
ratio  2;  required  the  13th  term. 

1,  2,     3,     4,     5,     6,     7,    indices. 

2,  4,     8,    16,  32,  64,  128,    leading  terms. 

Then  6+7=index  to  the  13th  term. 
And  64x123=8192  the  answer. 

2.  A  young  man  agreed  with  a  farmer  to  work  for  him 
11  years,  with  no  other  reward  than  the  produce   of  one 
grain  of  wheat  for  the  first  year,  allowing  the  increase  to 
he  tenfold,  and  that  produce  to  be  sowed  the  second  year, 
and  so  on  from  year  to  year,  until  the  end    of  the  time ; 
what  is  the   sum  of  the  whole    produce,    allowing  7680 
grains  to  make  a  pint,  and  what  does  it  amount  to  at  one 
dollar  and  fifty  cents  per  bushel  1 

Ans.  2^6056i-  +  busli.,  and  $339084, 19cts.+ 

NOTE. — In  such  questions,  you  first  find  the  last  term 
by  one  of  the  cases  in  Problem  2,  and  then  the  sum  of 
the  whole  series  by  Problem  1. 

3.  A  rich  miser  thought  20  guineas   apiece  too  much 
for  12  fine  horses,  but  readily  agreed  to  give  4  cents   for 
the  first,  16  cents  for  the  second,  64  cents  for  the  third 
horse,  and  so  on,  in  fourfold   proportion,  to  the   last ; — 
what  did  they  come  to  at  that  rate,   and  how   much   did 
they  cost  per  head,  one  with  another  ? 

The  \>  horses  came   to    S223f>96,20cts.,  and 
the  average  price  was  §18641,35cts.  per  head. 

CASE  2* — When  the  jirst  term  of  the  series ,  and  the  ra- 
tio >  are  different:  that  is,  when  the  first  term  is  either 
greater  or  less  than  the  ratio* 

RULE. —  1.  Write  down  a  few  of  the  leading  terms  of 
the  series,  aa  before,  and  begin  their  indices  with  a  ci- 
pher;  thus:  0,  1,  2,  3,  &c. 

*  When  the  first  term  of  the  series,  and  the  ratio  are  different, 
the  indices  must  beg-in  with  a  cipher,  and  the  sum  of  the  indices 
made  choice  of,  must  be  one  less  than  the  number  of  terms  given 
in  the  question  ;  because  1  in  the  indices  stands  orer  the  second 
term,  and  2  ui  the  indices  over  the  third  term,  &c. j  and.  ia  this; 

Q2 


186  GEOMETRICAL    PROGRESSION. 

2.  Add  together  the  most  convenient  indices  to  make 
an  index,  less  by  1,  than  the  number  expressing  the  place 
of  the  term  sought. 

3.  Multiply  the  terms  of  the  geometrical  series  togeth- 
er, belonging  to  those  indices,  and  make  the  product  a 
dividend. 

4.  Raise  the  first  term  to  a  power  whose  index  is  one 
less  than  the  number  of  terms  multiplied,  and  make  the 
result  a  divisor. 

5.  Divide  the  said  dividend  by  the   said  divisor,   and 
the  quotient  is  the  term  required. 

NOTE. — If  the  first  term  of  any  series  be  unity,  or  1, 
the  term  required  is  found  by  multiplying  the  terms  of 
the  geometrical  series  together  which  belong  to  those  in- 
dices, without  neecfing  any  division. 

EXAMPLES. 

1.  Required  the   12th  term  of  a  geometrical  series, 
whose  first  term  is  3,  and  ratio  2. 

0,     1,     2,     3,     4,     5,       0,  indices. 

3,     6,  12,  24,  48,  96,  192,  leading  terms. 

Then,  6+5=index  to  the  12th  term. 

And  192x96  =  l8432*=dividend. 

The  number  of  terms  multiplied  is  2,  and  2 — 1=1  is 
the  power  to  which  the  term  3  is  to  be  raised  ;  but  the 
first  power  of  3  is  3=divisor  ;  therefore 

18432—3=6144,  the  12th  term. 

2.  A  goldsmith  sold  Ijfo.  of  gold,  at  2cts.  for  the  first 
ounce,  Sets,  for  the  second,  32cts.  for  the  third,   &c.,  in 
quadruple  proportion  geometrical ;  what  did  the  whole 
come  to  1  Ans.  $111848,10cts. 

3.  A  man  bought  a  horse,  and  by  agreement  was  to  give 
a  farthing  for  the  first  nail,  two  for  the  second,   four   for 
the  third,  &c.     There  were  four  shoes,   and   eight  nails 
in  each  shoe  ; — What  did  the  horse  come  to  at  that  rate  ? 

Ans.  £4473924  5s.  3d.  3qrs. 

case,  the  product  of  any  two  terms,  divided  by  the  first  term,  is 
equal  to  that  term  beyond  the  first,  signified  by  the  sum  of  their 
indices.  rp,  JO,  1,  2,  3,  4,  £c.,  indices. 

:  \  1,     3,     9,  27,  81,  £c.,  geometrical  series. 
Here,  4+3=7,  the  index  of  the  8th  term. 
81x27=2187,  the  8th  term,  or  7th  beyond  the  1st, 


ALLIGATION    MEDIAL.  187 

4.  Suppose  a  certain  body,  put  in  motion,  should  move 
the  length  of  one  barleycorn  the  first  second  of  time,  one 
inch  the  second,  three  inches  the  third  second  of  time, 
and  so  continue  to  increase  its  motion  in  triple  proportion 
geometrical ;  how  many  yards  would  the  said  body  move 
in  the  space  of  half  a  minute  1 

Ans.  953199685623yds.  1ft.  lin.  Ibar.  j*  which  is  no 
less  than  five  hundred  and  forty-one  millions  of  miles. 


ALLIGATION. 

ALLIGATION  teaches  to  mix  several  simples  of  different 
qualities,  so  that  the  composition  may  be  of  a  middle 
quality  ;  and  is  commonly  distinguished  into  two  principal 
cases,  called  Alligation  Medial  and  Alligation  Alternate. 

ALLIGATION  MEDIAL. 

ALLIGATION  MEDIAL  is  the  method  of  finding  the  rate 
of  the  compound,  from  having  the  rates  and  quantities  ©f 
the  several  simples  given. 

RULE. — Multiply  each  quantity  by  its  rate  ;  then  divide 
the  sum  of  the  products  by  the  sum  of  the  quantities,  or 
the  whole  composition,  and  the  quotient  will  be  the  rate 
of  the  compound  required. 

EXAMPLES. 

L  Suppose  20  bushels  of  wheat  at  10s.  per  bushel,  36 
bushels  of  rye  at  6s.  per  bushel,  and  40  bushels  of  barley 
at  4s.  per  bushel,  were  mixed  together;   what  would   a 
bushel  of  this  mixture  be  worth  ? 
20x10=200 
36X  6=216 
40x  4=160 

96  )576(6s.  Answer. 

576 

2.  A  composition  being  made  of  5  pounds  of  tea  at  7s. 
per  pound,  9  pounds  at  8s.  6d.  per  pound,  and  14^- pounds 
at  6s.  lO^d.  per  pound  ;  what  is  a  pound  of  it  worth  ? 

Ans.  7s.  4f-d.+ 

3.  A  goldsmith  mixes  8  pounds  5£  ounces  of  gold  of 
14  carats  fine,  with  12  pounds  8£  ounces  of  18  ;  what  is 
the  fineness  of  this  mixture  ?  Ans.  IGy^y  carat*. 


188  ALLIGATION    ALTERNATE. 

4.  If  with  40  bushels  of  corn  at  4s.  per  bushel,  there 
are  mixed  10  bushels  at  6s.  per  bushel,  30  bushels  at  5s. 
per  bushel,  and  20  bushels  at  3s.  per  bushel  ;   what  will 
10  bushels  of  that  mixture  be  worth  ?     Ans.  $7,16§cts. 

5.  A  grocer  would  mix  12cwt.  of  sugar  at    10  dollars 
percwt.  with  3cwt.  at  8f  dollars  per  cwt.  and  8cwt.  at 
7£ dollars  per  cwt.  ;  what  will  a  cwt.  of  this  mixture  be 
worth  ?  Ans.  gS,95cts.  6mill«.+ 

6.  If  16  gallons  of  wine  at   1    dollar  25  cents,   and  4 
gallons  of  water,  be  mixed  with  40  gallons  of  wine   at  3 
dollars  per  gallon  ;  what  will  the  mixture  be   worth  per 
gallon  ?  Ans.  $2,33£cts. 

ALLIGATION  ALTERNATE. 

ALLIGATION  ALTERNATE  is  the  method  of  finding  what 
quantity  of  any  number  of  simples  whose  rates  are  given, 
will  compose  a  mixture  of  a  given  rate;  so  that  it  is  the 
reverse  of  alligation  medial,  and  may  be  proved  by  it. 

PCULE. — Write  the  rates  of  the  simples  in  a  column 
under  each  other. 

Connect,  or  link  with  a  continued  line,  the  rate  of  each 
.simple,  which  is  less  than  that  of  the  compound,  with  one 
or  any  number  of  those,  that  are  greater  than  the  com- 
pound ;  and  each  greater  rate  with  one  or  any  number 
of  the  If, 

Write  the  difference  between  the  mixture  rate,  and  that 
of  each  of  the  fim  pie,  Opposite  to  the  rates,  with  which 
they  are  respectively  linked. 

Then,  if  only  one  difference  stand  against  any  rate,  it 
will  be  the  quantity  belonging  to  that  rate;  but  if  there 
be  several,  their  sum  will  be  the  quantity. 

EXAMPLES. 

1.  A  merchant  would  mix  wines  at  14s.  15s.   19s.  and 

22s.  per  irall.  to  that  the  mixture  may  be  worth  18s.  per 
gallon  ;  what  quantity  of  each  must  be  taken  ? 

Or  thus  : 

1  ....  at  14s. 


Ans. 

J  3  ....  at  22s. 


ALLIGATIOK    ALTERNATE.  189 

2.  How  much  corn  at  2s.  6d.  3s.  8d.  4s.  and  4s.  8d.  per 
bushel,  must  be  mixed  together,  that  the  compound  may 
be  worth  3s.  lOd.  per  bushel  ? 

Ans.  12  at  2s.  6d.  12  at  3s.  8d.  18  at  4s,  &,  18  at  4s.  8d. 

3.  A  goldsmith  has  gold  of  18  carats  fine,  16,   19,  22 
and  24  ;  how  much  must  he  take  of  each  to  make  it  21 
carats  fine?    .        (  3oz.  of  16,  loz.  of   18,   loz.  of   19, 

3'  I  5oz.  of  22,  and  5oz.  of  24  carats  fine. 

4.  It  is  required  to  mix  wine  at  80  cents,  wine  at  70 
cents,  cider  at  lOcts.  and  water  together,  so  that  the  mix- 
ture may  be  worth  50cts.  per  gallon. 

Ans.  9gals.  of  each  sort  of  wine,  5  of  cider  &>  5  of  water. 

CASE  2. — When  the  whole  composition  is  limited  to  a  cer- 
tain quantity. 

RULE. — Find  an  answer  as  before  by  linking ;  then 
say,  as  the  sum  of  the  quantities,  or  differences  thus  de- 
termined is  to  the  given  quantity,  so  is  each  ingredient 
found,  to  the  required  quantity  of  each. 

EXAMPLES. 

1.  How  many  gallons  of  water  at  Octs.  per  gallon,  must 
be  mixed  with  wine  worth  60cts.  per  gallon  so  as  to  fill 
a  cask  of  100  gallons,  and  that  a  gallon  may  be  afforded 
at50cts.? 

*n/    °l10 
50  J  60  (50 

60  :  100  :  :  10  :         60  :  100  :  :  50  : 
10  50 


I 


6,0)100,0.  6,0)500,0 

16f  83* 

Ans.  16f  gallons  of  water,  and  83£  of  wine. 

2.  How  much  gold  of  15,  of  17,  of  18  and  22  carats 
fine,  must  be  mixed  together  to  form  a  composition  of  40 
ounces  of  20  carats  fine  ? 

Ans.  5oz.  of  15,  17  and  18,  and  25oz.  of  22. 

3.  Wine  at  3s.  6d.  and  at  5s.  9d.  per  gallon,  is  to  be 
mixed,  so  that  a  hogshead  of  63  gallons  may  be  sold  for 
<£12  12s. ;  how  many  gallons  must  be  taken  of  each  1 

Ans.  14gals.  at  5s.  9d.  and  49gals.  at  3s.  6d. 


190  SINGLE    POSITION. 

CASE  3.  —  When  one  of  the  ingredients  is  limited  to  a  cer- 
tain quantity. 

RULE.  —  Take  the  difference  between  each  price  and 
the  mean  rate  as  before  ;  then,  as  the  difference  of  that 
simple  whose  quantity  is  given,  is  to  the  rest  of  the  dif- 
ferences severally,  so  is  the  quantity  given  to  the  several 
quantities  required. 

EXAMPLES. 

1.  A  grocer  would  mix  teas  at  1  dollar  20cts.  66cts. 
and  1  dollar  per  pound,  with  20  pounds  at  40  cents  per 
pound  ;  how  much  of  each  sort  must  he  take  to  make  the 
composition  worth  80  cents  per  pound  ? 

ife    tfc     fb    tb 

40  :  20  :  :  20  :  10  at  66cts.  ) 
40:  14::  20:    7  at  $1        VAns. 
40:40:  :  20:  20  at     1,20  J 

low  much  wine  at  SOcts.  at  88  and  92  per  gallon, 
must  be  mixed  with  4  gallons  at  75cts.  per  gallon,  so 
that  the  mixture  may  be  worth  86  cents  per  gallon  ? 

Ans.  4gals.  at  SOcts.  8^  at  88  and  8J-  at  92. 
3.  With  95  gallons  of  wine  at  8s.  per  gallon,  I  mixed 
other  wine  at  6s.  8d.  per  gallon,  and  some  water  ;  then  I 
found  it  stood  me  in  6s.  4d.  per  gallon  ;  —  I  demand   how 
much  wine  at  6s.  8d.  I  took,  and  how  much  water. 
Ans.  95  gallons  wine  at  6s.  8d.  and  30  gallons  water. 


POSITION. 

POSITION  is  a  method  of  performing  such  questions  as 
cannot  be  resolved  by  the  common  direct  rules,  and  is  of 
two  kinds,  Single  and  Double. 

SINGLE  POSITION. 

Single  Position  teaches  to  resolve  those  questions 
whose  results  are  proportioned  to  their  suppositions. 

RULE.  —  1.  Take  any  number  and  perform  the  same 
operations  with  it,  as  are  described  to  be  performed  in 
the  question. 


DOUBLE    POSITION.  101 

2.  Then  say,  as  the  result  of  the  operation  is  to  the 
position,  so  is  the  result  in  the  question  to  the  number  re- 
quired. 

EXAMPLES. 

1.  A's  age  is  double  that  of  B,  and  B's  is  triple  that  of 
C,  and  the  sum  of  all  their  ages  is  140  ;  what  is  the  age 
of  each  ?  Suppose  A's  age  to  be  48 
Then  will  B's  =4£=24 
And  C's=-2^=8 

80  sum. 

As  80  :     48  ::     140     :     84  =  A's  age. 
Consequently  %4-=42=B's. 


140  Proof. 

2.  A  certain  sum  of  money  is  to  be  divided  between  4 
persons  in  such  a  manner  that  the  first  shall  have  ^  of  it, 
the  second  |,  the  third  £,  and  the  fourth  the  remainder, 
which  is  28  dollars  ;  what  is  the  sum  ?          Ans.  $112. 

3.  A  person,  after  spending  J-  and  £  of  his  money,  had 
60  dollars  left  ;  what  had  he  at  first  ?  Ans.  $144. 

4.  What  number  is  that  which  being  increased  by  £,  ^, 
and  £  of  itself,  the  sum  will  be  125  7  Ans.  60. 

5.  A  person  lent  his  friend  a  sum  of  money,  to  receive 
interest  for  the  same  at  6  per  cent,  per  annum,  simple 
interest  ;  at  the  end  of  three  years  he  received  for  princi- 
pal and  interest  383  dollars  50  cents  ;  what  was  the  sum 
lent  ?  Ans.  325  dollars. 

6.  A  cistern  is  supplied  with  three  cocks,  A,  B,  and  C  : 
A  can  fill  it  in  1  hour,  B  in  2,  and  C  in  3  ;  in  what  time 
will  it  be  filled  by  all  of  them  together  1  Ans.  T6T  hour. 


DOUBLE  POSITION. 

DOUBLE  POSITION  teaches  to  resolve  questions  by  mak- 
ing two  suppositions  of  false  numbers.* 

*  Questions  in  which  the  results  are  not  proportional  to  their 
positions,  belong1  to  this  rule  ;  such  are  those,  in  which  the  num- 
ber sought  is  increased  or  diminished  by  some  given  number, 
which  is  no  known  part  of  the  number  required. 


192  DOUBLE    POSITION. 

RULE. — 1.  Take  any  two  convenient  numbers,  and 
proceed  with  each  according  to  the  conditions  of  the 
question. 

2.  Find  how  much  the  results   are  different  from  the 
result  in  the  question. 

3.  Multiply  each  of  the  errours  by  the  contrary  suppo- 
sition. 

4.  If  the  errours  be  alike,  divide  the  difference  of  the 
products  by  the  difference  of  the  errours,  and  the  quo- 
tient will  be  the  answer. 

5.  If  the  errours  be  unlike,  divide  the  sum  of  the  pro- 
ducts by  the  sum  of  the  errours,  and  the  quotient  will  be 
the  answer. 

NOTE. — The  errours  are  said  to  be  alike,  when  they  are 
both  too  great,  or  both  too  little  ;  and  unlike,  when  one  is 
too  great,  and  the  other  too  little. 

EXAMPLES. 

1.  A  lady  bought  cambric  for  40  cents  a  yard,  and  In- 
dia cotton  at  20  .cents  a  yard  ;  the  whole  number  of  yards 
she  bought  was  8,  and  the  whole  cost  2   dollars  ;    how 
many  yards  had  she  of  each  sort  ? 

Suppose  4  yards  of  cambric,  value  $l,60cts. 
Then  she  must  have  4  yards  of  cotton,  value      80 

Sum  of  their  values,  2,40 
So  that  the  first  errour  is-f  40 

Again,  suppose  she  had  3  yards  of  cambric,  $l,20cts. 
Then  she  must  have  5  yards  of  India  cotton  1,00 

Sum  of  their  values,  2,20 
So  that  the  second  errour  is  +20 
Then  40 — 20 =20= difference  of  the  errours. 
Also  4x20=80=product  of  the  first  supposition  and 

second  errour. 
And  3x40=120=product  of  the  second  supposition 

and  first  errour. 

And  120— 80=40=their  difference. 
Whence  40^-20=2  yards  of  cambric.      >  ^^ 
And  8 — 2=6  yards  of  India  cotton.          / 

2.  A  and  B  have  both  the  same  income  ;  A  saves  £  of 
his  yearly  ;  but  B,  spending  50  dollars  a  year  more  than 


PERMUTATION   AND    COMBINATION.  193 

A,  at  the  end  of  4  years  is  100  dollars  in  debt ;  what  is 
their  income,  and  what  do  they  spend  per  annum  ? 

(  Their  income  is  $125  per  year. 
Ans.J  A  spends  $100. 

(B  spends  8150. 

3.  A  labourer  was  hired  for  40  days  upon  these  condi- 
tions, that  he  should  receive  2  dollars  for  every  day  he 
wrought,  and  forfeit  1  dollar  for  every  day  he  was  idle ; 
at  the  expiration  of  the  time  he  was  entitled  to  50  dollars  ; 
how  many  days  did  he  work,  and  how  many  was  he  idle  ; 

Ans.  He  wrought  30  days,  and  was  idle  10. 

4.  A  man  had  2  silver  cups  of  unequal  weight,  with  1 
cover  for  both,  weight  5oz  ;  now  if  he  put  the  cover  on 
the  less  cup,  it  will  be  double  the  weight  of  the  greater ; 
and  put  on  the  greater  cup,  it  will  be  three  times  the 
weight  of  the  less  cup ;  what  is  the  weight  of  each  cup? 

Ans.  3oz.  the  less,  and  4oz.  the  greater. 

5.  A   person   being   asked  what  o'clock  it  was,  an- 
swered that  the  time  past  from  noon  was  equal  to  -£%  of 
the  time  to  midnight  ;  required  the  time. 

Ans.  36  minutes  past  1. 

6.  There  is  a  fish  whose  head  is  ten  feet  long ;  his  tail 
is  as  long  as  his  head  and  half  the  length  of  his  body, 
and  his  body  is  as  long  as  his  head  and  tail  ;  what  is  the 
whole  length  of  the  fish  ?  Ans.  80  feet. 

7.  A  and  B  laid  out  equal  sums  of  money  in  trade  ;  A 
gained  a  sum  equal  to  £  of  his  stock,  and  B  lost  225  dol- 
lars ;  then  A's  money  was  double  that  of  B's ;  what  did 
each  lay  out.  Ans. 


PERMUTATION  AND  COMBINATION. 

THE  permutation  of  quantities  is  the  showing  how 
many  different  ways  the  order  or  position  of  any  given 
number  of  things  may  be  changed. 

The  combination  of  quantities  is  the  showing  how  of- 
ten a  less  number  of  things  may  be  taken  out  of  a  great- 
er, and  combined  together,  without  considering  their  pla- 
ces, or  the  order  in  which  they  stand. 
R 


Proof. 


194  PERMUTATION    AND    COMBINATION 

PROBLEM  1. 

To  find  the  number  of  permutations,  or  changes,  that 
can  be  made  of  any  number  of  things,  all  differing  from 
each  other. 

RULE. — Multiply  all  the  terms  of  the  natural  series  of 
numbers,  from  one  up  to  the  given  number,  continually 
together,  and  the  last  product  will  be  the  answer. 

EXAMPLES. 

1.  How  many  changes  may  be  made  with  these  three 
letters,  A,  B,  C. 

CHANGES. 

1  a  b  c 

2  a  c  b 
—                              b  a  c 

2  be  a 

3  cab 

c  b  a  6  J 
6  Answer. 

2.  How  many  changes  may  be  rung  upon  6  bells  ? 

Ans.  720. 

3.  How  many  changes  may  be  rung  upon  12  bells,  and 
how  long  would  they  be  ringing  but  once  over,  supposing 
10  changes  might  be  rung  in  one  minute,   and  that  the 
year  contains  365  days,  6  hours  1 

.        i  479001600  changes,  and  91 
5'  \  years  3w.  5d.  and  6  hours. 

4.  A  young  scholar  coming  into  a  town  for  the  conven- 
ience of  a  good  library,  demanded  of  the  gentleman  with 
whom  he  lodged,  what  his  diet  would  cost  for  a  year  ;  he 
told  him  $150;  but  the  scholar  not  being  certain  what 
time  he  should  stay,  asked  him  what  he  should  give  him 
for  so  long  as  he  could  place  his  family  (consisting  of  6 
persons  beside  himself)  in  different  positions  every  day  at 
dinner ;  the  gentleman  told  him  850 ;  to  this  the  scholar 
agreed — what  time  did  he  stay  1  Ans.  41)40  days. 

PROBLEM  II. 

Any  number  of  different  things  being  given,  to  find 
how  many  changes  can  be  made  out  of  them,  by  taking 
any  given  number  at  a  time. 


PERMUTATION    AND    COMBINATION.  195 

RULE. — Take  a  series  of  numbers,  beginning  at  the 
number  of  things  given,  and  decreasing  by  one,  till  the 
number  of  terms  be  equal  to  the  number  of  things  to  be 
taken  at  a  time,  multiply  these  terms  into  each  other  ; 
and  the  product  will  be  the  answer. 

EXAMPLES. 

1.  How  many  changes  may  be  made  out  of  the  three 
letters  a,  b,  c  by  taking  two  at  a  time  1 
CHANGES. 


3  a  b 

2  ba 

—  a  c 

6  Answer.  c  a 

be 

cb 


^  Proof. 


2.  How  many  changes  may  be  made  with  the  nine  di- 
gits, by  taking  3  at  a  time  ?  Ans.  504. 

3.  How  many  words  may  be  made  with  the  alphabet  by 
taking  five  letters  at  a  time,  supposing  that  a  number  of 
consonants  may  make  a  word  ?  Ans.  5100480. 

PROBLEM  III. 

To  find  the  compositions  of  any  number,  in  an  equal 
number  of  sets,  the  things  themselves  being  all  different. 

RULE. — Multiply  the  number  of  things  in  every  set 
continually  together,  and  the  product  will  be  the  answer, 

EXAMPLES. 

1.  Suppose  there  are  four  companies,  in  each  of  which 
there  are  nine  men;  it  is  required  to  find  how  many  ways 
four  men  may  be  chosen,  one  out  of  each  company. 

9x9x9x9=6561  the  answer. 

^.  How  many  changes  are  there  in  throwing  five  dice  ? 

Ans.  7776. 

3.  Suppose  there  are  four  companies,  in  one  of  which 
there  are  6  men,  in  another  8,  and  in  each   of  the  other 
two  9  ;  what  are  the  choices  by  a  composition   of  four 
men,  one  out  of  each  company  1  Ans.  3888. 

4.  Suppose  a  man  undertakes  to  throw   an  ace,  at  one 
throw,  with  4  dice  ;  what  is  the  probability  of  his  effect* 
ing  it  ?  Ans.  as  671  to  625, 


196  MISCELLANEOUS    QUESTIONS. 

MISCELLANEOUS   QUESTIONS. 

1.  A  gentleman  bought  27  yards  of  cloth  at  2s.  per 
yard,  24  yards  at  3s.  l^d.  per  yard,  25  yards  at  Is.  8£d. 
per  yard ;  he  also  bought  3  yards  of  broadcloth,  the  price 
of  which  he   does  not   recollect ;  but   on   counting  his 
money  he  found  he  had  expended  ^11   19s.  2£d. ;  what 
did  his  broadcloth  cost  per  yard,  in  Federal  Money  ? 

Ans.  $3,75cts. 

2.  A  servant  went  to  market  with  <£5,  and  bought  eggs 
at  7  for  4d. ;  2  pair  of  fowls  at  2s.  4d.  a  pair ;  17  pigeons 
at  3s.  per  dozen  ;  3  rabbits  at  I4d.  each;  and  3  dozen  of 
larks  at  14d.  per  dozen  ;  he  also  paid  the  baker  ,£2   17s. 
Id. ;  when  he  returned  he  had  21s.  left ;  how  many  eggs 
did  he  buy  ?  Ans.  126. 

3.  I  have  a  drawer  17  inches  long,  12  inches  broad  and 
7  inches  deep  ;  how  many  one  inch  dice  will  it  hold  1 

Ans.  1428. 

4.  At  a  certain  election  375  persons  voted,  and  the 
candidate  chosen  had  a  majority  of  91 :  how  many  voted 
for  each  ?  Ans.  233  and  142. 

5.  Suppose  a  man  to  step  30  inches  at  a  time,  and  to 
go  4  miles  an  hour  ;  how  many  times  does  he  step  in  a 
minute  ?  Ans.  140f . 

6.  The  divisor  is  43967,  the  quotient  2737226,  and  the 
remainder  27672 ;  what  is  the  dividend  ? 

Ans.  120347643214. 

7.  A  prize  of  $1000  is  to  be  divided  between  two  per- 
sons whose  shares  are  in  proportion  of  7  to  9 ;  required 
the  share  of  each.  A       (  S437,50cts. 

s*  \  g562,50cts. 

8.  After  paying  away  £  and  -^  of  my  money,  I  had  66 
guineas  left  in  my  purse  ;  what  was  in  it  at  first  ? 

Ans.  120. 

9.  A  reservoir  for  water  has  two  cocks  to  supply  it ; 
the  first  alone  will  fill  it  in  40  minutes,  the  second  in  50 
minutes  and  it  has  a  discharging  cock  by  which  it  may 
be  emptied,  when  full,   in  25  minutes.     Now  supposing 
that  these  three  cocks  are  all  opened,  that  the  water  comes 
in,  and  that  the  influx  and  the  efflux  of  the  water  are  al- 
ways alike.,  in  what  time  would  the  cistern  be  filled  ? 

Ans.  3  hours,  20  minutes* 


MISCELLANEOUS    QUESTIONS.  197 

10.  In  the  latitude  of  Hallowell,  a  degree  of  longitude 
measures  about  49  miles,  6  furlongs,  and  1 1^  poles  ;  now 
as  the  earth  turns  round  in  about  23  hours,  56  minutes, 
at  what  rate  per  hour  is  the  town  of  Hallowell  carried  by 
this  motion  from  west  to  east  1*       Ans.  748-i^J  miles. 

11.  If  the  earth  turns  round  in  23  hours  56  minutes,  at 
what  rate  per  hour  are  the  inhabitants  of  the  city  of  Quito 
in  South  America,  which  lies  under  the  equator,  carried 
from  west  to  east  by  this  rotation  1  Ans.  1045^1  miles. 

12.  In  a  mixture  of  wine  cider,  ^  of  the  whole  added 
to  25  gallons,  was  wine  ;  and  4-  part,  less  5  gallons,  was 
cider  ;  how  many  gallons  were  there  of  each  ? 

Ans.  85  of  wine  and  35  of  eider. 

1&.  A  hare  is  50  of  her  own  leaps  before  a  greyhound, 
and  takes  4  leaps  to  the  greyhound's  3  ;  but  two  of  the 
greyhound's  leaps  are  as  much  as  3  of  the  hare's  :  how 
many  leaps  must  the  greyhound  take  to  catch  tbe  hare  1 

Ans.  300. 

14.  Out  of  a  cask  of  wine  which  trad  leaked  away  J- 
part,  21  gallons  were  drawn  ;  and  then  'being  gauged,  it 
was  found  to  be  half  full ;  how  many  gallons  did  it  hold  t 

Ans,  126. 

15.  What  part  of  4d.  is  f  of  6  pence  ?  Ans.  f. 

16.  WJiat  number  is  that  from  which,  if  5  be  subtract- 
ed f  of  the  remainder  is  80  ?  Ans.  125. 

17.  A  post  is'J-  in  the  mud,  -J  in  the  water,  and  10  feet 
above  the  water  :  what  is  its  whole  length  ?      Ans.  24. 

18.  A  captain,  mate,  and  20  seamen,  took  a  prize  worth 
$3501  ;  of  which  the  captain   takes    11    shares,   and  the 
mate  5  shares  ;  the  remainder  of  the  prize  is  equally  di- 
vided among  the  sailors  ;   how  much   did   each   man   re- 
ceive ?      A       |      The    Capt.    $1069,75cts.— The  mate 


5,25cts.  Each  sailor  $97,25  cents. 

19.  A  stationer  sold  quills  at  $1,  83£cts.  per  thousand, 
by  which  he  cleared  |  of  the  money  ;  "but  as  they   grew 
scarcer,  he  raised  them  to    #2,25cts.    per  thousand  ; — 
what  did  he  clear  per  cent,  by  the  latter  price  ? 

Ans.  $96,36T4Tcts.  gained  per  cent,  by  the  latter  price, 

20.  Bought  a  quantity  of  goods  for  $250,  aud  3  month* 

*The  earth  moves  1  degree  in  3*  59*  2G**£f  of 
real  time,  or  in  4*  of  solar  or  tropical  time.. 
B2 


198  MISCELLANEOUS    QUESTIONS. 

after  sold  them  for  $275 ;  how  much  per  cent,  per  annum 
did  I  gain  by  them  ?  Ans.  $40. 

21.  In  what  time  will  the  interest  of  g72,60cts  equal 
that  of  $15,25cts.  for  04  days,  at  any  rate  of  interest  ? 

Ans.  13£f£  days. 

22.  What  sum  of  money  will  amount  to  $132,81cts. 
2jm.  in  15  months,  at  5  per  cent,  per  annum  simple  in- 
terest ?  Ans.  $125. 

23.  A  person  possessed  of  £  of  a  ship   sold  f  of  his 
share  for   $1260;  what   was  the  reputed  value  of  the 
whole  at  the  same  rate  ?  Ans.  5040. 

24.  Of  my  T%  of  a  farm  I  sell  f  of  f ;  what  I  then  own 
is  worth  8185  ; — what  is  that  part,  and  what  the  value  of 
the  farm  ?  Ans.  ^  and  lhe  farm  $1200. 

25.  What  number  is  that  to"which  if  T3^  of  4^-of  £f£  be 
added,  the  total  will  be  one  ?  Ans.  f f f f . 

26.  If  £  of  |  of  I  of  a  ship  be  worth  £  of  f  of  |£  of  the 
cargo,  valued  at  40000  dollars  ;  what  is  the  value  of  the 
ship  and  cargo  ?  Ans.  $50744,81  cts.+ 

27.  A  grocer  would  mix  a  quantity  of  sugar  at  lOd.  per 
pound,  with  other  sugars  at  7£d.  5d.  and  4£d.  per  pound, 
intending  to  make  up  a  compound  worth  6d.  per  pound  ; 
what  quantity  of  each  must  he  take?  Ans  l^ffo  at  lOd. 

Itb  at  7£d.  l£tb  at  5d.  and  4ft  at  4£d. 

28.  If  1000  men  besieged  in  a  town,with  provisions  for 
5  weeks,  allowing  each  man  16  ounces  a  day,  were  rein- 
forced with  500  men  more,  and  hear  that  they  cannot  be 
relieved  till  the  end  of  8  weeks  ;  kow  many  ounces  a  day 
must  each  man  have,  that  the  provisions  may  last  that 
time  ?  Ans.  6|  ounces. 

29.  Sound,  not  interrupted,  is  found  by  experiment  to 
move  uniformly  about  1150  feet  in  a  second  of  time  ;  how 
long  then,after  firing  an  alarm  gun  at  Fort  Independence* 
may  the  same  be  heard  at  Cambridge,  taking  the  distance 
at  5f  miles  ?  Ans.  26f  £  seconds. 

30.  If  I  see  the  flash  of  a  gun  fired  by  a  vessel  in  dis- 
tress at  sea,  which  happens,  we  will  suppose,  at  the  instant 
of  its  going  off,  and  hear  the  report  a  minute  and  3  sec- 
onds afterwards ;  how  far  is  she  off!  Ans.  72450  feet. 

SI*  An  elm  plank  is  14  feet,  3  inches  long ;  what  dis- 
tance from  the  edge  must  a  line  be  struck  to  take  off  a 
yard  square  1  Ans,  7|£  inches. 


MISCELLANEOUS    QUESTIONS.  199 

32.  A  man  dying  left  his  wife  in  expectation  that  a 
child  would  be  afterwards  added  to  the  family,  and  in 
making  his  will  ordered,  that  if  the  child  were  a  son,  f  of 
his  estate  should  belong  to  him,  and  the  remainder  to  his 
mother ;    but    if  it  were   a  daughter,  he  appointed  the 
mother  f  and  the  child  the  remainder  ;  but  it  happened 
that  the  addition  was  both  a  son  and  a  daughter,  by  which 
the  widow  lost  in  equity  2400  dollars  more  than  if  there 
had  been  only  a  girl ;  what  would  have  been  her  dowry, 
had  she  had  only  a  son  ?  Ans.  $2100. 

33.  Having  a  piece  of  land  11  perches  in   breadth,  I 
demand  what  length  of  it  must  be  taken  to  contain  an 
acre,  when  four  perches  in  breadth  require  40  perches  in 
length  to  contain  the  same  ?  Ans.  14per.  3yds. 

34.  If  a  gentleman  whose  annual  income  is  ^10009 
spends  20  guineas,  each  21s.,  a  week,  will  he  fall  in  debt, 
or  save  money,  and  how  much  in  the  year  ? 

Ans.  ,£92  in  debt. 

35.  What  sum  of  money  will  produce  as  much  interest 
in  3£  years,  as  $210,15cts.  can  produce  in  5  years  and  5 
months?  Ans.  $350,25cts. 

36.  If  $100  in  5  years  be  allowed  to  gain  820,  50cts. 
in  what  time  will  any  sum  of  money  double  itself,  at  the 
fame  rate  of  interest  ?  Ans.  24^f  years. 

37.  What  difference  is  there  between  the  interest  of 
•$350  at  4  per  cent,  for  8  years,  and  the  discount  of  the 
same  sum,  at  the  same  rate,  and  for  the  same  time  ? 

Ans.  $27,15^cts. 

38.  If  by  selling  goods  at  $2£  per  cwt.  I  gain  20   per 
cent.,  what  do  I  gain  or  lose  per  cent,  by  selling  at  $2£ 
per  cwt.  1  Ans.  $8  per  cent.  gain. 

39.  Required  the  length  of  a  shore,  which,  strutting 
11  feet  from  the  upright  of  a  building,  may  support  a 
jamb  23  feet,  9  inches  from  the  ground.  Ans.  26ft  2i».+ 

40.  A  clears  $13  in  6  months,  B  $18  in  5,  and  C  g23 
in  9,  his  stock  being  g72£ ;   what,  then,  is  the  general 
stock?  Ans.  $236,09£f. 

41.  A  person  making  his  wiff,  gave  to  one  child  £$  of 
his  estate,  and  the  rest  to  another  ;  when  these  legacies 
came  to  be  paid,  the  one  turned  out  $600  more  than  the 
other  :  what  did  the  testator  die  worth?     Ans.  $2000. 

42.  A  father  devised  -f^  of  his  estate  to  one  of  his  sons 
and  ^  of  the  residue  to  another,  and  the  remainder  to  bis 


200  MISCELLANEOUS    QUESTIONS. 

widow  for  life ;  the  children's  legacies  were  found  to  be 
S257,16f  cts.  different : — pray  what  sum  did  he  leave  the 
widow  the  use  of?  Ans.  $635,04T642rcts. 

43.  A  had  12  pipes  of  wine,  which  he  parted   with  to 
B  at  4^  per  cent,  profit,  who  sold   them  to   C  for   S40, 
GOcts.  advantage  ;  C  made  them  over  to  D  for  $605,50cts., 
and  cleared  thereby  6  per  cent.  : — how  much  a  gallon  did 
this  wine  cost  A  ?  Ans.    SS^^fyetB. 

44.  Laid  out  $I65,75cts.  in  wine  at  gleets,  a  gallon  ; 
some  of  which  receiving  damage  in  carriage,  I   sold  the 
rest  at  Slfcts.  a  gallon,  which  produced  only  $110,83^ 
cts.  ;  what  quantity  was  damaged  1  Ans.  430gals 

45.  A  young  hare  starts  40  yards  before  a    greyhound, 
and  is  not  perceived  by  him  till  she  has  been  up  40  sec- 
onds ;  she  scuds  away  at  the  rate  of  ten  miles  an  hour, 
and  the  dog,  on  view,  makes  after  her  at  the  rate  of  18  ; 
how  long  will  the  course  hold,  and  what  ground   will  be 
run  over  by  the  dog  ?     Ans.  60^-sec.  and  530  yds.  run. 

46.  If  I  leave  Hallowell  at  8  o  'clock  on  Monday  morn- 
ing for  Newburyport,  and  ride  at  the  rate  of  3  miles  an 
hour  without  intermission;  and  B  sets  out  from  Newbu- 
ryport for  Hallowell  at  4  o'clock  the  same  evening,  and 
rides  4  miles  an  hour  constantly:  supposing  the  distance 
between  the  two  towns  to  be    130  miles,  whereabout  on 
the  road  shall  we  meet  ? 

Ans.  69f  miles  from  Hallowell,  which  will  be  in  Saco. 

47.  X,  Y,  and  Z,  can,  working  together,  complete  a 
staircase  in  12  days  ;  Z  is  man  enough  to  do  it  alone  in 
24  days,  and  X,  in  34;  in  what  time,  then,  could   Y  get 
it  done  himself?  Ans.  8  If  days. 

48.  A  and  B  together  can  build  a  boat  in  18  days,  and 
with  the  assistance  of  C,  they  can  do  it  in    11   days;  in 
what  time  then,  would  C  do  it  by  himself? 

Ans.  28f  days. 

49.  Laid  out  in  a  lot  of  muslin  «^500^   upon  examina~ 
tion  of  which  3  parts   in   9  proved   damaged,   so  that  I 
could  make  but  5s.  a  yard  of  the  same  ;  and  by  so  doing, 
find  1  lost  =£50  by  it ;  at  what  rate  per  ell  Eng.  am  I  to 
part  with  the  undamaged  muslin,  in   order  to  gain  £50 
«pon  the  whole  ?  Ans.  11s.  7fd. 

60.  If  the  sun  move,  every  day,  one  degree,  and  the 
sxioon  thirteen ;  and,  at  a  certain  time,  the  sun  be  ai  the 


MEASUREMENT    OF    GRINDSTONES.  201 

beginning  of  Cancer,  and,  in  three  days  after,  the  moon 
at  the  beginning  of  Aries  ;  the  place  of  their  next  follow- 
ing conjunction  is  required.  Ans.  10°  45'  of  Cancer. 

51.  A  person  being  asked  the  time  of  day,   answered, 
it  is  between  4  and  5  ;  but  a  more  particular  answer  be- 
ing required,  he  said,  that  the  hour  and   minute  hands 
were  then  exactly  together  ;  what  was  the  time  7 

Ans.  21  T9Tmin.  past  4. 

52.  What  weight,  hung  at  70  inches'  distance  from  the 
fulcrum  of  a  steelyard,  will  equiponderate  a  hhd.  of   to- 
bacco, weighing  950&.  freely  suspended  at  2  inches'  dis- 
tance on  the  contrary  side!         Ans.  27ft.  2oz.  4^drs. 

53.  If  two  places  lie  so  much  due  east  and  west  of  each 
other,  that  it  is  found,  by  observation,  to  be  noon  at  the 
former  2  hours,  6  min.  and  30  seconds  sooner  than  at  the 
latter  ;  how  many  degrees  are  they  apart  1 

Ans.  31°  37'  30  seconds. 

54.  If  Paris,  in  France,  be  in  2°  20'  east  longitude 
from  Greenwich,  and  Hallowell  in  69°  42'  west  longitude 
from  Greenwich  ;  when  it  is  noon  at  Paris,  what  time  of 
day  is  it  at  Hallowell  ? 

Ans.  7h.  1  1m.  52s.  in  the  morning. 


MEASUREMENT  OF  GRINDSTONES. 

GRINDSTONES  are  sold  by  the  stone,  and  their  contents 
found  as  follows  ;* 

RULE.  —  To  the  whole  diameter  add  half  of  the  diam- 
eter, and  multiply  the  sum  of  these  by  the  same  half,  and 
this  product  by  the  thickness  ;  divide  this  last  number  by 
1728,  and  the  quotient  is  the  contents,  or  answer  re- 
quired. 

EXAMPLES. 

1.  What  are  the  contents  of  a  grindstone  24  inches 
diameter,  and  4  inches  thick  7 


24+12x12x4 

=1  stone.     Ans. 

1728 

2.  What  are  the  contents  of  a  grindstone  36  inches 
diameter,  and  4  inches  thick  :  Ans.  2£  stone. 

*  24  inches  in  diameter,  and  4  inches  thick,  make  a  stone. 


202  MENSURATION. 

MENSURA  TION  of  Superficies  and  Solids. 
Section  1. — OF  SUPERFICIES. 

Superficial  measure  is  that,  which  relates  to  length  and 
breadth  only,  not  regarding  thickness.  It  is  made  up  of 
squares,  either  greater  or  less,  according  to  the  different 
measures  by  which  the  dimensions  of  the  figure  are  taken 
or  measured.  Land  is  measured  by  this  measure,  its  di- 
mensions being  usually  taken  in  acres,  rods,  and  links. 
The  contents  of  boards,  also,  are  found  by  this  measure, 
their  dimensions  being  taken  in  feet  and  inches.  Be- 
cause 12  inches  in  length  make  1  foot  of  long  measure, 
therefore  12x12  =  144,  the  square  inches  in  a  superficial 
foot,  &c. 

CASE  1. —  To  find  the  Area  of  a  square  having  equal  sides. 
RULE. — Multiply  the  side  of  the  square  into  itself,  and 
the  product  will  be  the  area,  or  superficial  content,  of 
the  same  name  with  the  denomination  taken,  whether 
inches,  feet,  yards,  rods  and  links,  or  acres. 

EXAMPLES. 

1.  How  many  square  feet  of  boards  are  contained  in 
the  floor  of  a  room  which  is  20  feet  square  ? 

20x20=400  feet,  the  Answer. 

2.  Suppose  a  square  lot  of  land  measures  26  rods  on 
each  side,  how  many  acres  does  it  contain  1 

26x26 

As  160  square  rods  make  an  acre  :  therefore  —     — zi=4ac.  36  ro. 

160          Ans. 

CASE  2. —  To  measure  a  parallelogram  or  long  square. 

RULE. — Multiply  the  length  by  the  breadth,  and  the 
product  will  be  the  area,  or  superficial  content,  in  the 
same  name  as  that,  in  which  the  dimension  was  taken, 
whether  inches,  feet,  or  rods,  &c. 

EXAMPLES. 

1.  A  certain  garden,  in  form  of  a  long  square,  is  96  feet 
long,  and  54  feet  wide  ;  how  many  square  feet  of  ground 
are  contained  in  it?     96x54=5184  square  feet.  Ans. 

2.  A  lot  of  land,  in  form  of  a  long  square,  is  120  rods 
in  length,  and  60  rods  wide  ;  how  many  acres  are  in  it  ? 
120x60=7200  sq.  rods.  And  7200-i-160=45  acres.  Ans. 


OF    SUPERFICIES.  203 

3.  How  many  acres  are  in  a  field  of  oblong  form, 
whose  length  is  14,5  chains,  and  breadth  9,75  chains  ? 

Ans.  14ac.  Oroo.  22rods. 

NOTE. — The  Gunter's  chain  is  66  feet,  or  4  rods,  long, 
and  contains  100  links.  Therefore,  if  dimensions  be  giv- 
en in  chains  and  decimals,  point  off  from  the  product  one 
more  decimal  place  than  are  contained  in  both  factors 
and  it  will  be  acres  and  decimals  of  an  acre  ;  if  in  chains 
and  links,  do  the  same,  because  links  are  hundredths  of 
chains,  and,  therefore,  the  same  as  decimals  of  them. 
Or,  as  1  chain  wide,  and  10  chains  long,  or  10  square 
chains,  or  100000  square  links,  make  an  acre,  it  is  the 
same  as  if  you  divide  the  links  in  the  area  by  100000. 

4.  If  a  board  or  plank  be  21  feet  long,  and  18  inches 
broad,  how  many  square  feet  are  contained  in  it  ? 

18  inches  =1,5  foot.     And  21  x  1,5=3 1,5  feet.  Ans. 

Or,  in  measuring  boards,  you  may  multiply  the  length 
in  feet  by  the  breadth  in  inches,  and  divide  the  product 
by  12  ;  the  quotient  will  give  the  answer  in  square  feet, 
&c.  21x18 

Thus,  in  the  preceding  example, =31^- sq.  feet 

as  before.  12 

5.  If  a  board  be  8  inches  wide,   how  much  in  length 
will  make  a  foot  square  ? 

RULE.— Divide  144  by  the  width  ;  thus,  8)144 

Ans.     18  inch. 

6.  If  a  piece  of  land  be  5  rods  wide,  how  many  rods 
in  length  will  make  an  acre  1 

RULE. — Divide  160  by  the  width,  and  the  quotient  will 
be  the  length  required  ;  thus, 
5)160 

32  rods  in  length.  Ans. 

NOTE. — When  a  board,  or  any  other  surface,  is  wider 
at  one  end  than  the  other,  but  yet  is  of  a  true  taper, 
you  may  take  the  breadth  in  the  middle,  or  add  the 
width  of  both  ends  together,  and  halve  the  sum,  for  the 
mean  width  :  then  multiply  the  said  mean  breadth  in 
either  case,  by  the  length  ;  the  product  is  the  answer,  or 
area  sought. 


204  MENSURATION 

7.  How  many  square  feet  in  a  board  10  feet  long,  and 
13  inches  wide  at  one  end,  and  9  inches  wide  at  the 
other?  13+9 

•«=!!  inches  mean  width. 

2  ft.     in. 

10x11 

— =9|ft.  Ans. 

12 

8.  How  many  acres  are  in  a  lot  of  land  which  is  40 
rods  long,  and  30  rods  wide  at  one  end,  and  20  rods  wide 
at  the  other  1    30+20 

=25  rods  mean  width. 

2        Then,  25x40 

=61  acres.  Ans. 

160 

9.  If  a  farm  lie  250  rods  on  the  road,  and,  at  one  end, 
be  75  rods  wide,  and,  at  the  other,  55  rods  wide,  how 
many  acres  does  it  contain  ?      Ans.  10 lac.  2roo.  lOro. 

CASE  3. —  To  measure  the  surface  of  a  triangle. 

Definition. — A  triangle  is  any  three  cornered  figure 
which  is  bounded  by  three  right  lines.* 

RULE. — Multiply  the  base  of  the  given  triangle  into 
half  its  perpendicular  height,  or  half  the  base  into  the 
whole  perpendicular,  and  the  product  will  be  the  area. 

EXAMPLES. 

1.  Required  the  area  of  a  triangle  whose  base  or  longest 
side  is  32  inches,  and  the  perpendicular  height  14  inches. 
14~t~2=37=£  the  perpend,  and  32x7=224  sq.  in.     Ans. 

2.  There  is  a  triangular  or  three  cornered  lot  of  land, 
whose  base  or  longest  side  is  51  £  rods  ;  the  perpendicu- 
lar, from  the  corner  opposite  to  the  base,  measures  44 
rods  ;  how  many  acres  does  it  contain  ? 

44-r-2=22= ha  If  perpendicular. 
And  51,5x22 

=7  acres,  13  rods.  Ans. 

160 

*  A  triangle  may  be  either  right-angled  or  oblique ;  in  either 
case,  the  teacher  can  easily  give  the  scholar  a  just  idea  of  the 
base  and  perpendicular,  by  marking  it  down  on  a  slate  or  paper, 
&c.  In  a  right-angled  triangle,  the  longest  of  the  two  legs  which 
include  the  right-angle,  is  called  the  base ;  but  in  such  as  are 
oblique,  the  longest  of  the  three  sides  is  so  called. 


ON    SUPERFICIES. 


205 


3.  If  a  piece  of  land  lie  in  the  form  of  a  right-angled 
triangle,  its  base  being  37  rods,  and  the  perpendicular 
line  being  24£  rodfc,  how  many  acres  are  in  it  1 

Ans.  2,8617+  acres. 

4.  If  the  base  of  a  triangular  field  be  7  chains  and  50 
links,  and  the  perpendicular  4  chains  and  25  links,  how 
much  does  it  contain  ?  Ans.  lac.  2roo.  15rods. 

Joists  and  Planks  are  measured  by  the  following 
RULE. — Find  the  area  of  one  side  of  the  joist  or  plank, 
by  one  of  the  preceding  rules  ;  then  multiply  it  by  the 
thickness  in  inches  ;  and  the  last  product  will  be  the  su- 
perficial content. 

EXAMPLES. 

1.  What  is  the  area,  or  superficial  content,  or  board 
measure,  of  a  joist,  20  feet  long,  4  inches  wide,  and  3  in- 
ches thick?  20x4 

X3=20  feet.  Ans. 

12 

2.  If  a  plank  be  32  feet  long,  17in.  wide,  and  Sin.  thick, 
what  is  the  board  measure  of  it  ?  Ans.  136  feet. 

NOTE. — There  are  some  numbers,  the  sum  of  whose  squares 
makes  a  perfect  square ;  such  are  3  and  4,  the  sum  of  whose 
squares  is  25,  the  square  root  of  which  is  5  ;  consequently,  when 
one  leg-  of  a  right-angled  triangle  is  3,  and  the  other  4,  the  hy- 
potenuse must  be  5.  And  if  3,  4,  and  5,  be  multiplied  by  any 
other  numbers  each  by  the  same,  the  products  will  be  sides  of 
true  right-angled  triangles.  Multiplying  them  by  2,  gives  6,  8, 
and  10 — by  3,  gives  9,  12  and  15 — by  4,  gives  12,  16  and  20,  &c.  ; 
all  which  are  sides  of  right-angled  triangles.  Hence  architects 
in  setting  off  the  corners  of  buildings,  commonly  measure  6  feet 
on  one  side,  and  8  feet  on  the  other  ;  then,  laying1  a  10  feet  pole 
across  from  those  two  points,  it  makes  the  corner  a  true  right- 
angle. 

RULE  2. —  To  Jind  the,  area  of  any  triangle  when  the  three 

sides  only  are  given. 

RULE. — From  half  the  sum  of  the  three  sides  subtract 
each  side  severally  ;  multiply  these  three  remainders  and 
the  said  half  sum  continually  together ;  then  the  square 
root  of  the  last  product  will  be  the  area  of  the  triangle. 

EXAMPLE. 

Suppose  I  have  a  triangular  fish-pond,  whose  three  sides 
measure  400,  348,  and  312yds.  ;  what  quantity  of  ground 
does  it  cover  1  Ans.  10  acres,  3roods,  8-f-rods. 

S 


206  MENSURATION 

CASE  4.  —  To  measure  irregular  surfaces. 
RULE.  —  Divide  the  figure  or  plane  into  triangles,  by 
drawing  diagonal  lines  from  one  angle  to  another  ;  then 
measure  all  the  triangles,  by  either  of  the  rules  in  Case 
3  ;  and  the  sum  of  their  several  areas  will  be  the  area  of 
the  given  figure. 

EXAMPLE. 

If  a  piece  of  ground  be  divided  into  two  triangles  by 
a  diagonal  line  drawn  through  it  measuring  30  rods,  and 
two  perpendiculars  be  let  fall,  one  measuring  8  rods,  and 
the  other  14  rods  ;  how  many  acres  does  it  contain  ? 

Ans.  2*   acres. 


CASE  5.  —  To  measure  a  circle. 

Definition.  —  A  circle  is  a  figure  bounded  by  a  curve  or 
circular  line,  every  part  of  which  is  equally  distant  from 
the  middle  or  centre.  The  curve  line  is  called  the  per- 
iphery or  circumference  ;  a  line  drawn,  from  one  side  to 
the  other,  through  the  centre,  is  called  the  diameter  ;  and 
a  line  drawn,  from  the  centre  to  the  circumference,  is  call- 
ed the  semidiameter,  (half  diameter,)  or  radius. 

PROBLEM  I. 

The  diameter  given  to  Jind  the  circumference. 
RULE.  —  As  7  are  to  22,  so  is  the  given  diameter  to  the 
circumference  ;  or,  more  exactly,  as  113  are  to  355,  so  is 
the  diameter  to  the  circumference,  &c. 

EXAMPLES. 

1.  What  is  the  circumference  of  a  wheel  whose  diam- 
eter is  4  feet  ? 

As  7  :  22  :  :  4  :  12,57+feet  the  circumference.     Ans. 

2.  What  is  the  circumference  of  a  circle  whose  diam- 
eter is  35  rods  ?  As  7  :  22  :  :  35  :  110  rods.  Ans. 

NOTE.  —  To  find  the  diameter,  when  the  circumference 
is  given,  reverse  the  foregoing  rule,  and  say,  as  22  are  to  7, 
so  is  the  given  circumference  to  the  required  diameter  ;  or, 
as  855  are  to  11  3,  so  is  the  circumference  to  the  diameter. 

3.  What  is  the  diameter  of  a  circle  whose  circumfer- 
ence is  110  rodsl 

As  23  :  7  :  :  110  :  35  rods  the  diam.     An§. 


OF    SUPERFICIES.  207 

PROBLEM.  II. 
To  find  the  area  of  a  circle. 

RULE. — Multiply  half  the  diameter  by  half  the  circum- 
ference, and  the  product  is  the  area ;  or,  if  the  diameter 
alone  is  given,  multiply  the  square  of  the  diameter  by 
,785398  or,  which  is  near  enough,  by  ,7854 — and  the 
product  will  be  the  area. 

EXAMPLES. 

1.  Required  the  area,  or  superficial  content  of  a  circle 
whose  diameter>is  12  rods,  and  circumference  37,7  rods, 
18,85=half  the  circumference. 
6=half  the  diameter. 


113,10  area  in  square  rods.  Ans. 

2.  What  is  the  superficial  content  of  a  circular  garden 
whose  diameter  is  11  rods  I 

By  the  second  method. 

11x11=121.     ,7854x121=95,0334  rods.      Ans. 

3.  What  will  be  the  cost  of  a  circular  platform  to  the 
curb  of  a  round  well,  at  10^  cents  per  square  foot  ;  if  the 
diameter  of  the  well  be  42  inches,  and  the  breadth  of  the 
platform  be  14£  inches  1  Ans.  $l,S7^cts.+ 


PROBLEM  III. 

To  find  the  area  of  a  circle  when  the  circumference  alone  is  given. 

RULE.  —  Multiply  the  square  of  the  circumference  by 
,079577,  or,  which  is  near  enough,  by  ,07958  —  and  the 
product  will  he  the  area. 

NOTE.  —  ,785398  is  the  area  of  a  circle  whose  diameter 
is  1,  and  ,079577  is  the  area  of  a  circle  whose  circumfer- 
ence is  1. 

EXAMPLE. 

What  is  the  area  of  a  circle  whose  circumference  is  30 
rods?  30  x30x,  07958  =71,  62200  rods.  Ans. 

PROBLEM  IV. 

The  area  of  a  circle  given  to  Jind  the  diameter. 
RULE.  —  Divide  the  area  by  ,7854  —  and  the  square  root 
of  the  quotient  is  the  required  diameter* 


208 


MENSURATION 


EXAMPLES. 

1.  Required  the  diameter  of  a  circle  that,  will  contain 
within  its  circumference  the  quantity  of  an  acre  of  land. 
1  acre=^4840sq,  yds.     Then  1/,4-ff  J=88,5+yda.     Ans- 

2.  In  the  midst  of  a  meadow  abounding  with  feed, 
For  two  acres,  to  tether  my  horse,  I've  agreed  ; 
How  long  must  the  rope  be,  that,  feeding  all  round, 
He  may  n't  graze  less  or  more  than  the  two  acres  of 

ground  ?  Ans.  55^+yards. 

3.  A,  B,  and  C,  join  to  buy  a  grindstone,  36  inches  in 
diameter,  which  cost  $3£,  and  towards  which  A  paid  $1£, 
B,  $1^,  and  C,  83^cts.     The  waste  bole  for  the  spindle 
was  5  inches  square.     To  what  diameter  ought  the  stone 
to  be  worn,  when  B  and  C  begin  severally  to  work  with 
it  allowing  for  the  hole,  and  A  first  grinding  down  his 
share,  next  B,  and  then  C  ? 

j^       (  29,324+inch.  diameter  where  B  begins  to 

~\    grind;  and  19,013+in.  diam.  C  begins. 
NOTE. — Twice  the  square  of  the  side  of  a  square,  will 
bet  he  square  of  the  diameter  of  its  circumscribing  circle. 

PROBLEM  V. 

The  area  of  a  circle  given  to  jftnd  the  circumference. 
RULE. — Divide   the   given   area  by    ,07958 — and  the 
square  root  of  the  quotient  is  the  required  circumference. 

EXAMPLES. 

1.  The  expense  of  turfing  a  round  plot,  at  4  pence  per 
square  yard,  was  «£2.  9s.  9|d.  ,8 ;  what  was  its  circum- 
ference 1  Ans.  130+feet. 

2.  How  many  feet  of  boards  will  fence  a  round  garden, 
containing  just  two  acres,  the  fence  five  feet  high;  and 
what  will  be  the  expense  at  6£  mills  per  square  foot  ? 

Ans.  5231f +ft.  boards  ;  and  cost  $32,69cts.  6£m.+ 

CASE  6. —  To  measure  a  sector  of  a  circle. 

Definition. — A  sector  is  a  part  of  a  circle,  contained 
between  an  arch  line  and  two  radii,  or  semidiameters  of 
the  circle. 

RULE. — Find  the  length  of  the  arch  by  saying,  as  160 
degrees  are  to  the  number  of  degrees  in  the  arch,  so  is 
the  radius,  multiplied  by  3,1416,  to  the  length  of  the  arch, 
which  length,  divided  by  2,  and  multiplied  by  the  radius^ 
will  become  the  required  area. 


OF    SUPERFICIES.  200 

EXAMPLE. 

What  is  the  area  of  the  sector  of  a  circle  whose  radius 
ia  25  feet,  its  arch  containing  125°  ?* 
As  180°:  125°  ::   25x3,1416  :   54,5416+ft.  length  of 
arch.  54,5416 

Then, x25=681,77ft.  Ans. 

2 

RULE  2. — Find  the  area  of  a  circle  having  the  same 
radius ;  then  say,  as  360  degrees,  [the  number  of  de- 
grees into  which  all  circles  are  divided,]  are  to  the  area 
of  the  said  circle,  so  is  the  number  of  degrees  in  the  arch 
of  the  sector,  to  the  area  required. 
EXAMPLE. 

Required  the  area  of  a  sector  of  a  circle  whose  arch 
contains  65  degrees,  and  radius  35  feet.  Ans.  695^8 q.  ft* 
CASE  7. —  To  jind  the  area  of  a  segment  of  a  circle. 

Definition. — A  segment  of  a  circle  is  any  part  of  a 
circle  cut  off  by  a  right  line  drawn  across  the  circle, 
which  does  not  pass  through  the  centre,  and  is  always 
greater  or  less  than  a  semicircle. 

RULE. — Find  the  area  of  the  sector  having  the  same 
arch  as  the  segment,  by  Case  6  ;  find  also  the  area  of  the 
triangle  formed  by  the  chord  of  the  segment  and  the  radii 
of  the  sector,  by  Case  3  ;  subtract  the  area  of  the  latter 
from  that  of  the  former,  and  the  remainder  will  be  the 
area  of  the  segment,  when  the  segment  is  less  than  a 
semicircle :  but  the  sum  af  the  two  areas  is  the  answer* 
when  it  is  greater, 

EXAMPLE. 

What  is  the  area  of  the  segment,  whose  arch  contains- 
55°  ;  its  chord  12,5  rods;  the  perpendicular  of  its  trian- 
gle 16  rods  ;  and  its  semidiameter  17,2  rods-? 
First,  find  the  area  of  a  circle  whose  diameter  is  34,4  rods 
As     7  :  22  :  :  34,4  :  108,1  +  rods  circumference*. 
108,1  34,4 

X      =929,66  area  of  the  circle* 

2  2 

Then,  as  360°  :  929,66: :  55°  :  142  +  area  of  the  sector, 

*  As  we  have  not  been  able  to  obtain  engravings  to  represent 
any  of  the  figures  in  the  preceding  or  subsequent  Examples,,  the- 
Teacher  will,  we  trust,  be  so  good  as  to  draw  them,  for  the  PapiS, 
£>u  paper  or  a  slate* 

sa 


210  MENSURATION* 

And  the  chord  =  12,5  :  the  perpend.  16. 
12,5x16 
>        =100  rods,  area  of  the  triangle. 

2 

Area  of   the   sector=142 

Area  of  the  triangle  =  100 

Area  of  the  segment=42  rods.  Ans. 
NOTE. — A  regular  polygon,  whose  sides  and  angles  are 
all  equal,  may  be  measured  by  dividing  it  into  triangles, 
finding  the  area  of  one,  and  multiplying  this  area  by  the 
number  of  triangles  contained  in  the  polygon. 

CASE  8. —  To  describe,  and  Jind  the  areajsf  an  tllipse  or 

oval. 

RULE.— To  describe  an  ellipse  or  aval,  draw  aline>set 
one  foot  of  the  dividers  on  the  line,  as  a  centre,,  and  de- 
scribe a  circle  ;  move  the  dividers  to  some  other  point  on 
the  same  line,  [but  not  so  far  but  that  the  dividers  in 
forming  a  second  circle  may  extend  within  the  firsthand 
describe  a  second  circle  of  the  same  radius  as  the  former  ; 
then,  in  the  two  points  where  the  circles  intersect,  set  the 
dividers  to  complete  the  sides  of  the  oval ;  and  through 
these  intersecting  points  draw  the  line  called  the  conju- 
gate diameter^  crossing  the  line  first  drawn  called  the 
transverse  diameter,  in  the  centre  of  the  oval. 

RULE. — To  find  the  area  of  an  ellipse,  multiply  the 
transverse,  or  longest  diameter,  by  the  conjugate,  or 
shortest  diameter,  and  their  product  by  ,7854 :  and  the 
last  product  is  the  area  required* 

EXAMPLE. 

If  the  transverse  diameter  of  aa  oval  fish  pond  be  34 
iods>  and  the  conjugate  diameter  be  24  rods,  what  is  it* 
area  t  34  x24 X >7854  =640,8864  rods.  Ans. 

CASE  9* — To  Jind  the  area  of  a  globe  or  sphere. 

Definition. — A  sphere  or  globe  is  a  round  solid  body,  in 
the  middle  or  centre  of  which  is  an  imaginary  pointr 
from  whLeh  every  part  of  the  surface  is  equally  distant. 
An  apple,  or  a  ball  used  by  children  in  some  of  their 
pastimes,  may  be  called  a  sphere  or  globe. 

RuLE.-^-Multiply  the  circumference  by  the  diameter^ 
and  the  product  will  be  the  area,,  or  surface* 


OF    SOLIDS.  211 

EXAMPLES. 

1.  What  is  the  superficial  content  of  the  earth,  if  it  be 
360  degrees  in  circumference,  and  every  degree  measure 
69£  miles  ? 

360x69^=25020  circumf.  355  :  113  :  :  25020  :  7964  + 

diameter. 
And  25020x7964  =  199259280  area  in  squa.  miles.  Ans. 

2.  If  the  moon's  diameter  be  2180  miles,  what  is  her 
area  ?  Ans.  14928640+ square  miles. 

SECTION  II.— OF  SOLIDS. 

Solids  are  measured  by  the  solid  inch,  foot,  or  yard,  &c» 
1728  of  these  inches,  that  is  12x12x12,  make  1  cubic 
or  solid  foot. 

CASE  I. —  To  measure  a  Cube. 

Definition. — A  cube  is  a  solid  of  six  equal  sides,  each 
of  which  is  an  exact  square. 

RULE. — Multiply  the  side  by  itself,  and  that  product 
by  the  same  side,  and  this  last  product  will  be  the  solid 
content  of  the  cube. 

EXAMPLES. 

1.  If  the  side  of  the  cubic  block  be  18  inches,  or  1  foot 
and  6  inches,  how  many  solid  feet  does  it  contain? 

1ft.  6in.  =  1,5ft.  and  1,5x1,5x1,5=3,375  solid  ft.  Ans. 

in.    in.    in.. 
Or,  18x18x18 

=3,375  as  before. 

1728 

2.  Suppose  a  cellar  is  to  be  dug  which  shall  contain  1£ 
feet  every  way,  in  length,  breadth,  and  depth  ;  how  many 
solid  feet  of  earth  must  be  taken  out  to  complete  it. 

Ans.  1728  sol.  ft. 

CASE  2. —  To  find  the  content  of  any  regular  solid,  of 
three  dimensions,  length,  breadth,  and  thickness,  suck 
as  a  piece  of  square  timber,  urhose  length  is  more  than 
its  breadth  and  depth. 

RULE. — Multiply  the  breadth  by  the  depth  or  thick- 
ness, and  that  product  by  the  length  ;  the  last  product  i* 
the  solid  content. 


212  MENSURATION 

EXAMPLES. 

1.  How  nany  solid  feet  are  there  in  a  piece  of  square 
timber  that  is  1  foot  and  6  inches,  or  ISiuches  Abroad,  9 
inches  thick,  and  9  feet,  or  108  inches  long  ? 

1ft.  Gin.  =  1,5  foot.     ,75x1,5x9=10,125  sol.  ft.  Ans. 
9  inches =,75  foot, 
in.    in.    in. 
Or,  18x9x108 

=10,125  as  before. 

1728 

In  measuring  timber,  however,  you  may  multiply  the 
breadth  in  inches  by  the  depth  in  inches,  and  that  pro- 
duct by  the  length  in  feet :  divide  this  last  product  by  144, 
and  the  quotient  will  be  the  solid  content  in  feet,  &c. 

2.  How  many  solid  feet  does  a  piece  of  square  timber, 
or  a  block  of  marble,  contain,  if  it  be  16  inches  broad, 
11  inches  thick,  and  20  feet  long  ? 

16x11x20=3520,  and  3520— 144  =24,4+ sol.  ft.  Ans. 

3.  If  a  stick  of  square  timber  be   15  inches  broad,  8 
inches  thick,  and  25  feet  long,  how  many  solid   feet  are 
in  it?  Ans.  20,8  + feet. 

CASE  3. — When  the  breadth  and  thickness  of  a  piece  of 
square  timber  arc  given  in  inches,   to  Jind  how  much   in 
length  will  make  a  solid  foot. 
RULE. — Divide  1728  by  the  product  of  the  breadth  and 

depth,  and  the  quotient  will  be  the  length,  making  a  solid 

foot. 

EXAMPLES. 

1.  In  a  piece  of  square  timber  11  inches  broad  and  8 
inches  deep,  what  length  will  make  a  solid  foot  1 

Ilx8=88)l728(19,6+inches.  Ans. 

2.  In  a  piece  of  square  timber  18  inches  broad  and  14 
inches  deep,  what  length  will  make  a  solid  foot  ? 

Ans.  6,8+inches. 

CASE  4. —  To  measure  a  cylinder. 

Definition. — A  cylinder  is  a  round  body  whose  bases  or 
ends  are  circles,  like  a  round  column  or  stick  of  timber, 
of  equal  bigness  from  end  to  end. 

RULE. — Multiply  the  square  of  the  diameter  of  the 
base  or  end  by  ,7854,  which  will  give  the  area  of  the 
base  ;  then  multiply  the  area  of  the  base  by  the  length,, 
and  the  product  will  be  the  solid  content* 


OF    SOLIDS.  213 

EXAMPLES. 

1.  What  is  the  solid  content  of  a  round  stick  of  timber, 
or  a  marble  column,  of  equal  bigness  from   end  to  end, 
whose  diameter  is  18  inches,  and  length  20  feet  ? 

1 8  inches  = 1,5ft.  l,5x  1,5  x, 7854  =  1,767 15  area  of  the 
base.     1,76715x20  length  =35,343  solid  feet.  Ans. 
Or,  18 x  18  x, 7854  =254,4696  inches,  area  of  the  base. 

254,4696x20 

=35,347  as'before. 

144 

2.  What  is  the  solid  content  of  a  round  stick  of  tim- 
ber, of  equal  bigness  from  end  to  end,  whose  diameter  is 
35  inches,  and  length  35  feet  1  Ans.  233,842  feet. 

CASE  5. —  Tojind  how  many  solid  feet  a  round  stick  of 
timber,  equally  thick  from  end  to  end,  will  contain,  when 
hewn  square. 

RULE. — Multiply  twice  the  square  of  its  semidiameter, 
in  inches,  by  the  length  in  feet ;  then  divide  the  product 
by  144,  and  the  quotient  will  be  the  answer. 

EXAMPLES. 

1.  If  the  diameter  of  a  round  stick  of  timber  be   22 
inches,  and  its  length  20  feet,  how  many  solid  feet  will  it 
contain  when  hewn  square  ? 

11x11x2x20 

Half  diameter =11,  and —  ==33,6-f  ft.   the 

144 
solidity  when  hewn  square,  the  answer. 

2.  If  the  diameter  of  a  round  stick  of  timber  be  24 
inches  from  end  to  end,  and  its  length  20  feet,  how  many 
solid  feet  will  it  contain,  when  hewn  square,  and  what 
will  be   the   content   of  the   slabs  which   reduce  it  to  a 
square  ?          *       j  40  feet  solidity  when  hewn  square, 

s<  \  arid  22,832ft.  the  solidity  of  the  slabs, 

CASE  6. —  To  Jind  how  many  feet  of  square  edged  boards^ 
of  a  given  thickness,  can  be  sawn  from  a  log  of  a  giv- 
en diameter. 

RULE. — Find  the  solid  content  of  the  log,  when  made 
square,  by  the  last  Case  ;  then  say,  as  the  thickness  of 
the  board,  including  the  saw  calf,  is  to  the  solid  feet,  so 
are  12  inches  to  the  number  of  feet  of  boards. 

*       -W 


214  MENSURATION 

EXAMPLES. 

1.  How  many  feet   of  square  edged  boards,   l£  inch 
thick,  including  the  saw  calf,  can  be  sawn  from  a  log  20 
feet  long,  and  24  inches  diameter  7 

12x12x2x20 

=40ft.  solid  content  when  hewn  square. 

144 

As  1£  :  40  :  :  12  :  384  feet.  Ans. 

2.  How  many  feet  of   square  edged  boards,   1£  inch 
thick,  including  the  saw  gap,  can  be  sawn  from  a  log  12 
feet  long,  and  18  inches  diameter]  Ans.  108  feet. 

NOTE. — A  short  rule  for  finding  a  number  of  feet  of 
one  inch  boards  that  a  log  will  make,  is  to  deduct  £  of  its 
diameter  in  inches,  and  ^  of  its  length  in  feet ;  then  for 
each  inch  of  diameter  that  remains,  reckon  1  board  of 
the  same  width  as  this  reduced  diameter,  and  of  the  same 
length  as  this  reduced  length  of  the  log  :  thus  a  log  12 
feet  long,  and  12  inches  through,  gives  9  boards,  9  feet 
long,  9  inches  wide,  or  60f  feet — a  log  16  feet  long,  and 
16  inches  through,  gives  12  boards,  12  inches  wide,  12 
feet  long,  or  144  feet. 

CASE  2. —  The  length,  breadth,  and  depth  of  any  cubical 
box  being  given,  to  Jind  how  many  bushels  it  will  con- 
tain. 

RULE. — Multiply  the  length,  breadth  and  depth  togeth- 
er, in  inches,  and  divide  the  last  product  by  2150,425,  the 
solid  inches  in  the  statute  bushel,  and  the  quotient  will 
be  the  answer. 

EXAMPLE. 

There  is  a  square  or  cubical  box  ;  the  length  of  its  bot- 
tom is  50  inches,  breadth  of  ditto  40  inches,  and  its  depth 
60  inches  ;  how  many  bushels  of  corn  will  it  hold  ? 
50x40x60 

=55,8+  or  55bush.  3  pecks.  Ans. 

2150,425 

CASE  8. —  To  Jind  the   solidity    of  a   cone    or  pyramid, 

whether  round,  square,  or  triangular. 
Definition. — Solids  which  decrease  gradually  from  the 
base  till  they  come  to  a  point,  are  generally  called  cones 
or  pyramids,  and  are  of  various  kinds,  according  to  the 


or  SOLIDS.  215 

figure  of  their  bases  ;  round,  square,  oblong,  triangular, 
&c.  ;  the  point  at  the  top  is  called  the  vertex,  and  a  line 
drawn  from  the  vertex,  perpendicular  to  the  base,  is  call- 
ed the  height  of  the  pyramid. 

RULE. — Find  the  area  of  the  base,  whether  round 
square,  oblong,  or  triangular,  by  some  one  of  the  forego- 
ing rules,  as  the  case  may  be  ;  then  multiply  this  area  by 
one-third  of  the  height,  and  the  product  will  be  the  solid 
content  of  the  pyramid. 

EXAMPLES. 

1.  What  is  the  solid  content  of  a  true-tapered  round 
stick  of  timber,  24  feet  perpendicular  length,  15  inches 
diameter  at  one  end,  and  a  point  at  the  other? 

15xl5x,7S54X8 

=9,8 175  solid  feet.  Ans. 

144 

2.  What  is  the  solid  content  of  a  square  stick  of  timber 
of  a  true  taper,  30  feet  perpendicular  length,   18  inches 
square  at  one  end,  and  a  point  at  the  other  1  Ans.  22£feet. 

3.  What  is  the  solid  content  of  a  triangular  tapering 
stick  of  timber,  21  feet  long,  10  inches  each  side  of  the 
triangle,  8f  inches  the  perpendicular  of  the  triangle  at 
the  large  end,  and  the  other  end  a  point  ? 

Half  perpendicular=4,33  and  4,33x10x7 

=2, 1  ft. + Ans. 

144 

NOTE. — If  a  stick  of  timber  be  hewn  three  square,  and 
be  equal  from  end  to  end,  you  find  the  area  of  the  base  as 
in  the  last  question,  in  inches,  multiply  that  area  by  the 
whole  length,  and  divide  the  product  by  144,  to  obtain 
the  solid  content. 

4.  If  a  stick  of  timber  be  hewn  three  square,  be  12  feet 
long,  and  each  side  of  the  base  10  inches,  the  perpendic- 
ular of  the  base  being  8f  inches,  what  is  its  solidity  ? 

Ans.  3,6+ feet. 

CASED. —  To  Jind  the  solidity  of  the  frustum  of  a  eone 

or  pyramid. 

Definition. —  The  frustum  of  a  cone  is  what  remains 
after  the  top  is  cut  off  by  a  plane  parallel  to  the  base,  and 
is  in  the  form  of  a  log  greater  at  one  end  than  theothera 
whether  round,  or  hewn  three  or  four  square,  &c. 


216  MENSURATION 

RULE. — If  it  be  the  frustum  of  a  square  pyramid,  mul- 
tiply the  side  of  the  greater  base  by  the  side  of  the  less ;  to 
this  product  add  one  third  of  the  square  of  the  difference 
of  the  sides,  and  the  sum  will  be  the  mean  area  between 
the  bases ;  then  multiply  this  sum  by  the  height,  and  it 
will  give  the  content  of  the  frustum.  Or,  if  it  be  a  taper- 
ing square  stick  of  timber,  take  the  girth  of  it  in  the  mid- 
dle ;  square  £  of  the  girth,  (or  multiply  it  by  itself  in  inch- 
es ;)  then  say,  as  144  inches  to  that  product,  so  is  the 
length,  taken  in  feet,  to  the  content  in  feet. 

EXAMPLE. 

What  is  the  content  of  a  tapering  square  stick  of  tim- 
ber, whose  side  of  the  largest  end  is  12  inches,  of  the 
least  end,  8  inches,  and  whose  length  is  thirty  feet,  calcu- 
lating it  by  both  rules  ?  4x4 

By  the  first  Rule :     12x8=*96.       12—8=4 =5£ 

_  3 

And  96+5^x30 

=2l£ft.  Ans. 

144 
By  the  second  Rule  :  12+8 

=10in.=£  of  the  girth  in 

the  middle.  2 

Then  10xlO=100=area  in  the  middle  of  the  stick. 
And,  as  144  :  100  :  :  30ft.  :  20,  83+ feet.     Ans. 

RULE. — If  it  be  a  triangle  pyramid,  or  a  tapering 
three  square  stick  of  timber,  multiply  the  sum  of  the 
mean  area,  as  found  in  the  first  rule,  by  ,433— and  that 
product  by  the  height  or  length.  Or,  multiply  the  area 
in  the  middle,  as  found  in  the  second  rule,  by  ,433 — and 
then  state  the  proportion  as  before. 
EXAMPLE. 

What  is  the  content  of  a  tapering  three-square  stick  of 
timber,  whose  side  of  the  largest  end  is  15  inches,  of  the 
least  end,  6  inches,  and  whose  length  is  40  feet,  calcu- 
lating it  by  both  rules  ?  9x9 

By  the  first  Rule:  15x6=90.     15—6=9. =27. 

3 

And90+27x,433x40 

«=  14,0725ft.  Ans. 

144 


OF    SOLIDS.  217 

15+6 

By  the  second  Rule  : =10,5  in.=£  of  the  girth 

2 

in  the  middle,  if  it  were  four-square. 
Then    10,5 x  10,5 x, 433 =47,73825in.-=area   in    middle. 
And,  as  144:  47,73825  :  :  40  feet  :  13,260625ft.  Ans, 

RULE. — If  it  be  a  circular  pyramid  or  cone,  multiply  the 
diameters  of  the  two  bases  together,and  to  the  product  add 
one  third  of  the  square  of  the  difference  of  the  diameters  ; 
then  multiply  this  sum  by  ,7854 — and  it  will  be  the  mean 
area  between  the  two  bases ;  multiply  this  area  by  the 
length  of  the  frustum,  and  it  will  give  the  solid  content. 

Or  multiply  each  diameter  into  itself;  multiply  one 
diameter  by  the  other ;  multiply  the  sum  of  these  pro- 
ducts by  the  length  ;  annex  two  ciphers  to  the  product, 
and  divide  it  by  382 ;  the  quotient  will  be  the  content, 
which  divide  by  144  for  feet  as  in  other  cases. 

EXAMPLE. 

What  is  the  solid  content  of  a  tapering  round  stick  of 
timber,  whose  greatest  diameter  is  13  inches,  the  least 
6£  inches,  and  whose  length  is  24  feet,  calculating  it  by 
both  rules  ? 

13x6,5=84,5       13—6,5=6,5      6,5x6,5 

=14,083+ 

3 

And  84,5+1 4,083  x, 7854x24 

= 12,904 + feet.    Ans. 

144 
By  the  second  Rule  ;* 


13x13+6,5x6,5+13x6,5x2400 

=1858,1 15+ 

382 
And  1858,115~144  =  12,903+ft.  Ans. 

*  To  find  the  content  of  timber  in  the  tree,  multiply  the  square 
of  1-5  of  the  circumference  at  the  middle  of  a  tree,  in  inches,  by 
twice  the  length  in  feet,  and  the  product  divided  by  144  will  be 
the  content,  extremely  near  the  truth.  In  oak  an  allowance  of 
1-10  or  1-12  must  be  made  for  the  bark,  if  on  the  tree ;  in  other 
wood  less*  Trees  of  irregular  growth  must  be  measured  in  parts. 
T 


218  MENSURATION    OF    SOLIDS. 

CASE  10. —  To  Jind  the  solid  content  of  a  Sphere  or  Globe. 

NOTE. — For  definition  of  a  Globe,  see  Case  9  of  Su- 
perficies. 

RULE. — Find  the  superficial  content  by  Case  9  of  Su- 
perficies;  multiply  this  surface  by  one-sixth  of  the  diam- 
eter, and  it  will  give  the  solidity. 

Or,  multiply  the  cube  of  the  diameter  by  ,5236 — and 
the  product  will  be  the  solidity.* 

EXAMPLE. 

What  is  the  solidity  of  our  earth,  if  its  diameter  be 
7957  f  miles,  nearly,  and  its  circumference  at  the  equator 
be  just  25000  miles? 

7957,75x25000x7957,75-f-6=~263857106187,5-f  solid 

miles.     Ans. 

CASE  11. —  To  Jind  the  solid  content  of  a  frustum  or  seg- 
ment of  a  Globe. 

Definition. — The  frustum  of  a  globe  is  any  part  cut 
off  by  a  plane. 

RULE. — To  three  times  the  square  of  the  semidiameter 
of  the  base,  add  the  square  of  the  height ;  multiply  this 
sum  by  the  height,  and  the  product  again  by  ,5236;  the 
last  product  will  be  the  solid  content. 

EXAMPLE. 

If  the  height  of  a  coal-pit,  at  the  chimney,  be  9  feet, 
and  the  diameter  at  the  bottom  be  24  feet,  how  many 
cords  of  wood  does  it  contain,  allowing  nothing  for  the 
chimney  ? 

24-r-2=12=semidiam.      12x12x3=432.     9x9=81. 

And  432+81  x9x,5236 

=18,886+ cords.  Ans. 

128=solid  feet  in  a  cord. 

*If  the  diameter  of  a  sphere  be  1,  its  solidity  will  be  ,5236; 
and  if  its  circumference  be  1,  its  solidity  will  be  ,016887. 


GAUGING.  219 

SECTION  III. 

OF  CASK  GAUGING. 

Definition. — Gauging  is  the  finding  of  the  content  of 
any  Cask,  Box,  Tub,  or  other  Vessel. 

Among  the  many  different  rules  for  gauging,  the  fol- 
lowing is  as  exact  as  any. 

RULE. — Take  the  diameter  at  the  hung  and  head,  and 
length  of  the  cask  ;  subtract  the  head  diameter  from  the 
bung  diameter,  and  note  the  "difference. 

If  the  staves  of  the  cask  be  much  curved  or  bulging 
between  the  bung  and  head,  multiply  the  difference  of  di- 
ameters by  ,7 ;  if  not  quite  so  much  curved,  by  ,65 ;  if 
they  bulge  yet  less,  by  ,6  ;  and  if  they  are  almost  or  quite 
straight,  by  ,55 — and  add  the  product  to  the  head  diame- 
ter ;  the  sum  will  be  a  mean  diameter. 

Square  the  mean  diameter,  thus  fou-nd,  and  multiply 
the  square  by  the  length  ;  divide  the  product  by  359  for 
ale  or  beer  gallons,  and  by  294  for  wine  gallons.  _ 

NOTE. — 1.  To  measure  the  length  of  the  cask,  take 
the  length  of  the  stave  ;  then  take  the  depth  of  the  chimes, 
which,  with  the  thickness  of  the  heads,  (that  are  1  inch, 
lvj-  inch,  or  2  inches,  according  to  the  size  of  the  cask) 
being  subtracted  from  the  length  of  the  stave,  leaves  the 
length  within. 

2.  In  taking  the  bung  diameter  observe  by  moving  the 
rod  backward  and  forward  whether  the  stave  opposite  to 
the  bung,  be  thicker  or  thiner  than  the  rest,   and  if  it 
be,  make  allowance  accordingly. 
EXAMPLE. 

How  many  ale  and  wine  gallons  will  a  cask  contain, 
whose  bung  diameter  is  30  inches,  head  diameter  25  in- 
ches and  length  40  inches  ? 

30—25=5.     5x,7=3,5  25  +  3,5=28,5  mean 

diam. 

28,5x28,5x40 

..-=90,5+  ale  gal.     Ans. 


359 


28,5x28,5x40 

=110,51 +wine  gal.     Ans. 

294 


220  GAUGING. 

Or,  by  the  sliding  rule.  On  D.  is  18,94— the  gauge- 
point  for  ale  or  beer  gallons,  marked  A.  G  :  and  17,14— 
the  guage-point  for  wine  galons,  marked  W.  G.  Set  the 
gauge-point  to  the  length  of  the  cask  on  C.  and  against 
the  mean  diameter,  on  D.  you  will  have  the  answer  in  ale 
or  wine  gallons,  accordingly  as  which  gauge-point  you 
make  use  of.* 

CASE  2. —  To  gauge  round  tubs,  fyc. 
RULE. — Multiply  one  diameter  by  the  other,  and  to 
that  product  add  one  third  of  the  square  of  their   differ- 
ence ;  multiply  this  sum  by  the  length,  and  divide  as  be- 
fore for  beer  or  wine. 

EXAMPLES. 

What  is  the  content,  in  beer  and  wine  gallons,  of  a 
round  tub,  whose  diameter  at  the  top,  within,  is  40  inches, 
and  at  the  bottom  34  inches,  and  the  perpendicular  height 

36  inches  ?  

6x6  34x40+12x36 

=12. =137£+beer  gal. 

40— £4=6    3  359  Ans. 


And  34x40+ 12x36 

=168  wine  gal.     Ans. 

294 

CASE  3. —  To  gauge  a  square  vessel. 
RULE. — Multiply  the  length  by  the  breadth,  and  that 
product  by  the  depth  ;  then  divide  by  282  for  beer  or  ale, 
(the  inches  in  a  beer  or  ale  gallon,)and  by  231  for  wine, 
&c.,  (the  inches  contained  in  a  wine  gallon,)  and  the 
quotient  will  be  the  answer. 

*  A  rule  which  has  been  given  as  generally  more  exact,  is  this  » 
multiply  the  product  of  the  square  of  the  mean  diameter  and  the 
length,  by  34,  and  point  off  four  places  from  the  right  of  the  pro- 
duct ;  the  figures  on  the  left  of  the  point  will  be  the  gallons,  and 
those  on  the  right  decimal  parts  of  a  gallon,  in  wine  or  cider* 
Let  the  dimension  be  taken  exactly  in  inches  and  tenths.  Take 
the  preceding-  Example  in  Case  1. 

30+25 

=27,5  mean  diam. 

2 

and  27,5x27,5x40x34=  1 028500,00 =102^gall on s  of 
wine  or  cider  ^  which  i&7y%6<jgal.  less  than  by  the  other.. 


SHIPb'    TONNAGE,    &C.  221 

EXAMPLE. 

If  a  square  vessel  he  80  inches  in  length,  60  in  breadth, 
and  40  inches  deep,  what  is  its  content  in  beer  and  wine 
gallons  ? 
80x60x40     wine  gal.  80x60x40     beer  gal. 

. -63l,l(3+Ans. =680,85+ Ans, 

231  282 

NOTE. — The  content  of  any  vessel,  in  feet,  gallons, 
and  bushels,  may  be  thus  found  :  Measure  the  inside  of 
the  vessel,  according  to  the  rule  of  the  figure,  and  find 
the  content  in  cubic  inches  ;  then, 

C  1728  }  and  the        C  Cubic  feet. 

TV   •  i    i      j     282  f  quotient      y  Ale  or  beer  gal. 

5  b^  )     231  f  will  be  the  ^  Wine  gallons. 

\.  2150,425    )  content  in  V.  Bushels. 

SECTION  IV. —  To  find  the  tonnage  of  a  ship. 

RULE. — Multiply  the  length  of  the  keel  by  the  breadth 

of  the  beam,  and  that  product  by    the   depth  of  the   hold  ; 

divide  the  last  product   by   95,   arid    the   quotient    is  the 

tonnage.  If  double  decked,  half  the  breadth  is  the  depth. 

EXAMPLE. 

If  a  ship  be  72  feet  by  the  keel,  24  feet  by  the  beam, 
and  12  feet  deep,  what  is  the  tonnage  1 


Ans. 

Or,  RULE  2. — Multiply  the  length  of  the  keel  by  the 
breadth  of  the  beam,  and  that  product  again  by  half  the 
breadth  of,  the  beam  ;  divide  the  last  product  by  94,  and 
the  quotient  is  the  tonnage.* 

EXAMPLE. 

What  is  the  tonnage  of  a  ship  that  is  84  feet  by  the 
keel,  arid  28  feet  by  the  beam  ?  23-f-2  =  14 


84x28x1 4^94=350,29  -f  tons.     Ans, 
NOTE. — The  breadth  of  the  beam  added  to  two    thirds 
the  length  of  the  keel,  gives  the  length  of  a  ship's  main-* 

*  Rule  established  by  Congress.  For  double  decked  vessels ;  length 
frora  fore  part  of  main  stem  to  ~afterpart  of  stern  post,  above  tipper  deck  ; 
breadth  at  videst  part,  above  main  wales,  La.f  of  which  is  called  the  depth  .;, 
deduct  from  length  3-5ths  of  breadth  3  multiply  the  remainder 


222    STRENGTH    OF    CABLES. SHIPS*    BURTHEN,  &C. 

mast :  Therefore  the  length  of  the  mainmast  of  the  ship 
last  mentioned,  is  84x2-f-3+28=84  feet. 

To  find  the  thickness,  the  proportion  is,  as  84  to  28,  so 
is  the  length  of  a  mast  in  feet,  to  its  thickness  in  inches. 
Consequently  the  thickness  of  the  mast  whose  length  was 
just  found,  is  28  inches. 

SECTION  V* — TfrfindtJie  weight  of  anchors  which  cables 
may  sustain. 

RULE. — As  the  strength  of  cables,  and  consequently 
the  weights  of  their  anchors,  are  proportioned  to  the 
cubes  of  their  peripheries ;  therefore,  as  the  cube  of  the 
periphery  of  any  cable,  is  to  the  weight  of  its  anchor,  so 
is  the  cube  of  the  periphery  of  any  other  cable,  to  the 
weight  of  its  anchor. 

EXAMPLES. 

1.  If  a  cable  of  6  inches  round  require  an  anchor  of 
2£cwt.  what  would  be  the  weight  of  an  anchor,  for  a  12 
inch  cable  ?  cwt. 

6x6x6    :  2,25  ::  12x12x12  : 18cwt.  Ans. 

2.  If  a  12  inch  cable  required  an  anchor  of  18cwt.  what 
must  the  circumference  of  a  cable  be,  for  an  anchor  of 
2£cwt.  T  cwt>  cwt. 

18  ;  12x12x12  :  :  2,25:  216.  ^216=6in.  Ans. 

SECTION  VI. — From  one  solid's  capacity,  to  find  another's. 

RULE. — As  the  cube   of  any  dimension  is  to  its  given 

weight*  so  is  the  cube  of  any  like  dimensions  to  its  weight. 

EXAMPLES. 

1.  If  a  ship  of  300  tons'  burthen  be  75  feet  by  the 
keel,  what  is  the  burthen  of  one,.  100  feet  by  the  keel,  of 
tike  fornv,  25|fc.  a  qiv  ? 

tons.  tons.  cwt.  \fo* 

75 3  : 300  i :  1003  s  71 1    2  22f .  Ans. 

2.  If  »  brass  cannon,  lljinch.  diameter,  weigh  lOOOffe. 
what  will  another,  20»8&  inch  diameter,,  of  like  metal 
and  shape,  weigh  *  Ans.  5942,5697ft.  + 

and  the  product  by  depth  ;  divide  by  95;  the  quotient  is  the  true  tonnage. 
For  single  decked,  take  depth  from  under  side  of  deck  plauk  to  ceiling  ia 
t&e  bolci;,  and  proceed  as  before^ 


WOOD    MEASURE.  -  ASSESSMENT    OF   TAXES.        223 

SECTION  VII. 
OF  WOOD  AND  BARK  MEASURE. 

CASE  1.  —  To  find  the  solid  content  of  wood  4*  bark. 
RULE.  —  Multiply  the  length,  breadth,  and  thickness  to- 
gether, agreeably  to  the  rule  of  Duodecimals,  and  the  last 
product  will  be  the  solid  content  of  the  pile,  parcel,  or  load. 

EXAMPLE. 

If  a  load  of  wood  be  8  feet  4  inches  long,  3  feet  8  in- 
ches wide,  and  4  feet  6  inches  high,  how  many  cubic  feet 
does  it  contain  ?  ft.  /  ft.  y  ft.  t 

8  4x3  8x4  6=137£  sol.  ft.  Ans. 
CASE  2.  —  To  Jind  how  many  cords  of  wood  or  bark  are 

contained  in  any  pile,  4*c. 

RULE.  —  Find  the  solid  content  as  before,  and  divide 
that  product  by  128  ;  the  quotient  will  be  the  cords,  and 
the  remainder  cubic  feet,  or  so  many  128ths  of  a  cord. 
Or,  divide  the  solid  content  of  the  pile,  &c.  by  16,  and 
the  quotient  will  be  cord-feet,  8  of  them  being  1  cord, 
and  the  remainder  so  many  16ths  of  a  cord-foot. 
EXAMPLES. 

1.  In  a  pile  or  load  of  wood  9  feet  4   inches  long,  3 
feet  8  inches  wide,  and  4  feet  9  inches  high,  how  many 
cords,  and  how  many  cord-feet  ? 

ft-,  ft-  ,  ft-  <   *olft.iM  ft.  „ 

94x38x49=16268  And  162  6  8-f-  128=1  cord, 

and  34  sol.  ft.  6  8.    Ans. 

Or,    162  68-4-16=10|cord  ft.  6  8=1  cord  2J-  feet.  Ans. 

2.  If  a  load  of  wood  or  bark  be  8  feet  long,  4  feet  wide, 
and  2  feet  6  inches  high,  how  many  cord-feet  does  it  con* 
tain  ?  Ans.  5  feet,  or  £  of  a  cord. 


MODE  OF  ASSESSING  TAXES. 

It  may  not  perhaps,  be  here  amiss  to  show  the  general  method  of 
Assessing-  Taxes.  But  as  the  quantity  of  new  matter  with 
which  we  have  enlarged  this  edition  of  our  Work,  has  extend- 
ed its  pages  considerably  beyond  the  limits  first  intended,  a 
brief  explanation  of  the  general  principle  and  rule,  will,  we 
trust*  fully  suffice  far  the  purpose. 


224  ASSESSMENT    OF    TAXES. 

ARGUMENT. 

There  is  a  certain  town  which  contains  8  inhabitants,  whom  we  will  call 
A,  B,  C,  D.  E,  F,  G,  and  H.  The  town  is  divided  into  2  school  districts  or 
classes,  which  are  numbered  1  and  2.  A,  B,  C,  and  D,  form  District  INo.  1, 
and  E,  F,  G,  and  H,  No.  2.  On  these  inhabitants  the  following1  taxes  are  to 
be  assessed,  namely  : 

State,  $14,  88cts.  6m. 

County,  19,  84cts.  8m. 

Town,  39,  69cts.  6m. 

School,  29,  77cls.  2m. 


fixed  ;  and  for  that  purpose  all  assessors  consult,  of  course,  the  latest  acts  that 
have  been  passed  on  the  subject. 

Let  a  poll  be  taxed  $1,30  ;  an  acre  of  orchard  25cts. ;  an  acre  of  tillage 
16cts.  5  an  acre  of  mowing  IGcts.  5  an  acre  of  pasturage  4-cte.  $.  an  ox  of  five 
years  35cts.  ;  and  a  cow  20cts. 

In  order  to  find  each  person's  proportion  of  the  several  taxes,  and  each 
school  district's  proportion  of  school  money,  according  to  the  rateable  estates 
of  the  members  of  each  district  or  class,  or  according  to  the  number  of  schol- 
ars in  each  district ;  each  man's  inventory  must  be  taken,  and  the  amount 
cast  by  the  following  rule. 

RULE. — Multiply  the  value  of  a  poll  by  his  number  of  polls ;  his  acres  of 
orchard  by  the  tax- value  of  one  ;  his  number  of  oxen  by  the  tax-rate  of  one  j 
and  so  of  every  other  kind  of  property  ;  add  the  products,  and  the  sum  is  the 
amount  of  his  rateable  estate;  find  the  amount  of  all  in  the  same  way  ;  add 
these  amounts,  and  their  sum  is  the  value  of  the  inventory  of  the  town. 

I  demand  the  rateable  estate  of  A,  who  has 
2  Polls,  at  $1,30  amount  to  $2,60 

2  Acres  of  tillage,  at     0,16  0,32 

5  Acres  of  mowing",  at     0^16  0,80 

2  Oxen,  at     0,35  0,70 

Amount  of  A*s  rateable  estate,  $4,42 

Find  the  other  amounts,  in  the  following*  inventory,  in  the 
same  way.  / 

To  prove  the  Inventory. 

RULE. — Add  up  the  column  of  polls,  and  multiply  the  sum  by 
the  value  of  one:  add  up  each  of  the  other  columns,  and  multi- 
ply its  sum  by  the  tax-rate  of  one  in  that  column;  then  add  the 
several  amounts  of  the  columns  tog-ether,  and  the  sum  will  be 
equal  to  the  total  amount  of  the  fnventory,  if  the  work  be  rig-lit. 

EXAMPLE. 

The  total  amount  of  rateable  estates  in  the  following*  invento- 
ry, is  $49,  62cts. 

"  And  proceeding  by  the  method  given  in  the  rule  of  proof,  the 
sum  of  the  products  is  $49,  62cts,  It  is,  therefore,  evident  the 
work  is  right. 

*  Assess  money  taxes  so  far  over  the  sum  to  be  raised,  as  to  meet  abate- 
ments. 


ASSESSMENT    OF    TAXES. 


225 


0 

1 

c 

£ 

3 

•a 
"o 

§2 

§3 

NOTE.  —  If  any  teacher  think  it 

o 

"S 

P 

best  to  proportion  the  School  Money 

O  ^ 

P 

§ 

as 

&H 

s-i 

<$  W 

between  the  districts,  according  to  the 

GO 

d 

iO 

^5 

number  of  scholars  in  each,  instead  of 

Q 

CO 

c»  .a 

Q    O 

0 

o 

"o 

G 

«   S 

-*-i    fw 

by  the  value  of  rateable  estates  in 

rt 

£  B 

y 

6 

Q 

Q 

X 

> 

Q 

o  « 

each,  let  the  scholar  do  it  so  ;  and  let 

£ 

< 

O 

o 

district  No.  1  contain  15,  and  district 

A 

B 

2 

1 

3 

2 

5 

5 
2 

4 

2 

1 

$4,42 
3,53 

No.  2,  20  scholars.    The  inventory 
here  given,  though  it  exhibits  but  a 
few    rateable  articles,  will  serve  to 

C 
D 

4 
I 

6 

3 

8 

10 

2 

2 
1 

9,96 
1,50 

explain  the  principle.     As  minors  now 
pay  no  poll-tax  in  Maine,  no  person 

E 

3 

10 

5 

8 

2 

7,02 

can  properly,  have  more  than  1  poll  ; 

F   1 

5 

4    6 

5 

2 

1 

5,25 

though  he  may  pay  the  tax  for  his 

G 

2 

4 

8 

4 

6 

2 

6,46 

workmen,  and  his  sons  who  are  of 

H 

5 

12 

8 

12 

2 

3 

11,48 

age. 

Total  Amt.  $49,62 

To  find  each  person's  proportion  of  any  tax. 

RULE. — Say,  as  the  total  amount  of  the  Inventory  of  the 
town,  is  to  the  sum  to  be  raised  in  each  tax,  so  is  1  dollar  to  that 
part  of  the  tax  which  one  dollar  of  the  Inventory,  or  rateable 
estate,  must  pay :  then,  taking-  the  same  numbers  for  the  first 
and  second  terms,  and  one  cent  for  the  third  term,  of  a  new 
stating-,  find  what  part  of  the  tax  one  cent  of  the  Inventory,  or 
rateable  estate,  must  pay ;  and  from  these  two  operations  form 
two  tax  tables ;  one  for  dollars,  from  1  dollar  to  11,  or  farther,  if 
deemed  necessary  ;  and  the  other  for  cents,  from  one  cent  to 
90.  Then  by  means  of  these  two  tables,  make  out  each  per- 
son's tax. 

1.  To  make  the  State  tax,  the  sum  to  be  raised  being*  $14,88cts. 
6m.,  and  the  total  amount  of  the  foregoing-  Inventory  $49,62cts. 

As  $49,62cts.  :    $14,  88cts.  6m.     :  :     $1,  OOcts.  :  $0,  30cts. 

And  as  $49,  62cts.  :  $14,  88cts.  6m.   :  :  ,01ct.  :   ,00cts.   3m. 

Therefore,  $1  of  the  Inventory  pays  30  cents  ;  and  1  cent  of 
the  Inventory  pays  3  mills ;  by  which  make  the  following-  two 
Tables. 


DOLLAR  TABLE, 

from  $1  to  $11. 

$  $  cts. 

I  pays  0,  30 


CENT  TABLE, 

from  1  Cent  to  90  Cents. 


2 

3 

4 

5 

6 

7 

8 

9 

10 

11 


0,  60 

0,  90 

1,  20 
1,  50 

1,  80 

2,  10 
2,  40 

2,  70 

3,  00 
3,  30 


cts. 

cts. 

m. 

cts. 

cts. 

1  pays 

0 

3 

30  pays 

9 

2 

0 

6 

40 

12 

3 

0 

9 

50 

15 

4 

1 

2 

60 

18 

5 

1 

5    '     70 

21 

6 

1 

8   '     80 

24 

7 

2 

1 

90 

27 

8 

2 

4 

9 

2 

7 

10 

3 

20 

6 

i 

226  BOOK-KEEPING. 

The  tax  is  now  to  be  made  on  each  rateable  estate,  as  it  stands 
in  the  Inventory,  by  means  of  these  tables.  —  First,  What  js  A's 
tax,  whose  rateable  estate  is  $4.  42cts.  ? 
By  the  table,  $4  pay  $1,  20cts. 

40cts  pay          0,12 
and  2  "     ^  0,00,  6m. 

Amount  $1,  32,  6  A's  tax. 

Or,  having-  found  what  part  of  the  tax  one  cent  of  the  Invento- 
ry will  pay,  you  may,  instead  of   making-  tables,  multiply  the 
number  of  cents,  in  each  person's  Inventory,  by  what  one   cent 
pays  and  the  product  will  be  his  tax. 
Now,  to  find  A's  tax  by  this  method  : 
One  cent  pays  3  mills,  or  ,3  of  a  cent. 
Therefore,  442cts. 
,3 

132,6=132TV5ts.  or  $1,  32cts.  6m.  as  before. 

Find  by  these  methods,  the  State  tax  of  all  the  other  persons. 
Then,  to  know  if  your  work  be  rig-lit,  add  the  several  persons' 
taxes  tog-ether,  'and  see  if  the  sum  be  just  equal  to  the  $14  88cts. 
6m.  that  was  raised  for  the  State,  which  it  must  be,  because  the 
proportion  is  even. 

Next,  find  each  person's  County  tax  in  like  manner,  taking  new 
stating-s,  and  forming-  new  tables  :  and  thus  proceed  with  each 
particular  tax,  till  you  have  gone  through  the  whole,  proving 
each  part  as  before  noted. 

Lastly,  form  your  tax  list,  setting-  down  the  names  therein  al- 
phabetically, and  carrying-  out  in  a  line  from  each  the  separate 
sums  of  the  respective  taxes,  together  with  the  total  amount  of 
each.  When  done,  give  them  a  general  proof,  by  adding  togeth- 
er the  several  sums  that  were  to  be  raised  for  State,  County,  &c. 
taxes  :  and  then  the  total  amounts  of  each  person's  taxes;  which 
two  sums  will  come  exactly  alike,  if  there  be  no  errour,  in  any 
part  of  the  work. 


BOOK-KEEPING. 


DIRECTIONS  FOR   THE  LEARNER. 

Having-  ruled  your  books  in  the  proper  form,  copy  inlo  the  Daybook  one 
day's  accounts;  then  calculate  them  upon  your  slate  or  waste-paper,  to 
find  if  they  be  rightly  cast  up,  and  to  exercise  you  in  calculations.  Next, 
rule  your  slate  or  waste-paper  in  the  form  of  the  Leger,  and  upon  it  post 
the  accounts  that  you  have  copied  in  the  Daybook,  with  their  date  prefix- 
ed 5  observing  to  set  on  the  Dr.  side  of  each  person's  account,  those  ac- 
counts to  which  he  is  Dr.  in  the  Daybook,  and  on  the  O.  side  of  his  ac- 
count, those  by  which  he  is  Cr.  And  if  any  account  consist  of  but  one 
article,  you  are  to  express  it  particularly  with  its  amount,  in  the  columns; 
but  if  it  consits;  of  several  articles,  write  To  or  By  Sundries,  placing  the 

JI^L 


BOOK-KEEPING. 


227 


sum  of  the  amounts  of  all  the  articles  in  the  columns.  After  the  accounts 
are,  by  correcting  if  necessary,  placed  according  to  the  teacher's  mind, 
transcribe  them  into  your  Leger,  leaving  a  proper  space,  under  each  per- 
son's name,  to  receive  more  accounts.  Then  under  the  proper  letters  in 
the  Alphabet,  enter  those  names  with  the  pages  where  they  stand  in  the 
Leger  5  and,  lastly,  write  the  Daybook  pages  to  the  several  accounts  in 
the  Leger,  by  which  you  can  readily  refer  to  the  page  of  the  Daybook  on 
which  any  Leger  entry  may  be  found,  making  at  the  same  time,  the  marks 
on  the  Daybook  which  denote  the  several  accounts  to  be  posted.  Do  the 
same  with  the  next  day's  accounts  :  and  so  on  till  the  whole  be  finished. 
But  observe  that  you  must  not  enter  any  person's  name  down  again  which 
has  been  entered  before,  till  the  space  first  assigned  to  it  shall  be  filled  with 
articles  ;  and  then  the  account  must  be  tranferred  to  a  new  place,  as  you 
may  observe  is  done  with  George  Sampson's  account. 
EXAMPLE. 

Suppose  David  Davis  owes  me  450  dollars  for  the  balance  of  an  account 
with  him,  April  1st.  1822;  the  next  day,  April  2d.  I  buy  of  him  200  bush- 
els of  wheat  at  1  dollar  50  cents  per  bushel,  and  100  bushels  of  corn  at  75 
cents  per  bushel  ;  the  next  day,  April  3d,  I  sell  Jonathan  Worth  150  bush- 
els of  wheat'  at  1  dollar  75  cents  per  bushel  j  April  4th,  Jonathan  Worth 
pays  me  200  dollars  in  cash,  and  David  Davis  pays  me  50  dollars  in  cash  : 
'required  the  Daybook  and  Leger  of  the  transaction. 


DAYBOOK,  NO.  1. 


Hallowdl,  April  1,  1822. 


= 

David  Davis. 
To  balance  due  on  old  account, 
/?«**•«'*> 

Dr. 

$ 
450 

cts 

00 
00 

00 
50 
00 

= 

David  Davis, 
By  20  bushels  wheat,      . 
100     do.     corn, 

April  3 

Or. 

at  $1,  50 
75 

300 
75 

375 



Jonathan  Worth, 
To  150  bushels  wheat, 

•    •                                       /?r»T*i/   A. 

Dr. 

at  $1,75 

Cr. 

262 

200 

Jonathan  Worth, 
By  cash  in  part  for  wheat, 

David  Davis, 
By  cash,  fifty  dollars, 

Cr. 

50 

00 

To  post  the  above  accounts,  open  an  account  for  David  Davis,  debit  him 
for  450  dollars;  and  for  the  second  day's  transaction  credit  him  for  375 
dollars  :  for  the  third  open  an  account  for  Jonathan  Worth,  debiting  him 
for  262  dollars  50  cents  ;  and  for  the  fourth  day  credit  him  for  200  dollars, 
and  credit  David  Davis  for  50  dollars. 


228 


L  EGER  NO.  1. 
1822.        Dr.  David  Davis, 


April  1 
April  4 

To  balance  of  old  account, 
To  balance  of  above  account, 

227 

$| 
450 

450 

Cts. 
00 

00 

25 

00 

1822. 

Dr.         Jonathan  Worth, 

April  3 

i 

To  wheat, 

227 

262 

50 

262 

50 

4 

To  balance  of  old  account, 

62 

50 

By  the  above  Leger  it  appears  that  the  balances  are  in 
my  favour,  which,  added  to  the  cash  I  have  on  hand,  and 
the  goods  unsold,  show  the  amount  of  my  stock,  which 
compared  with  my  original  stock,  will  show  my  profit  or 


loss,  viz. 


$.  Cts. 
25  00 
62  50 
25000 


David  Davis  owes  me 
Jonathan  worth  do. 
I  have  in  cash, 
Wheat  unsold  50  bushels, 

valued  at  prime  cost,  $1,50 
Corn  do.  100  bushels  do.  at  75  cts.  75  00 


7500 


Amount  of  my  stock, 
My  original  stock  wai 


I  have  therefore  gained 


487  50 
450  00 

37  50 


(1*) 

LEGER,  NO.  I. 


S29 


1822.         Contra 


Cr. 


April  2 
4 

By  Sundries,         - 
By  Cash, 
By  balance  to  new  account, 

227 

$ 
375 
50 
25 

Cts. 
00 
00 
00 

00 

450 

1822. 

April  4 

Contra                       Cr. 

By  cash,         -                 - 
By  balance  to  new  account, 

227 

200 
62 

00 
50 

50 

262 

NOTE. — If  you  should  enter  any  thing  in  your  Leger 
under  a  wrong  title,  or  in  any  other  way  false,  it  should 
not  be  blotted  out,  but  marked  thus  (x)  in  the  margin 
against  it ;  and  write  on  the  opposite  side  Err  our  per  con- 
tra, with  the  sum  against  it,  and  make  the  same  mark  in 
the  margin. 

*  Opposite  pages  in  the  Leger  are  both  numbered  alike. 


230 


0) 
DAYBOOK,  No.  2. 


Hallotoell,  April  G,  1827. 


George  Simpson 
To  balance  due  on  old  account, 


Dr. 


John  Barton  Dr. 

To  balance  due  on  account  of  6  ) 
Hhds.  Wine.  } 


William  Reed 


Cr. 


By  balance  due  him  on  account   of  > 
English  goods  pr.  invoice,  ) 


Thomas  Tilton  Cr. 

By  balance  due  him  for  6  months 
service  on  farm, 

April  8. 


Charles  Prince  Dr. 

To  1  2  I7£tb-  sugar  25fca  qr.  at  $12 
2  bbls.  superfine  flour,  at  7    50 

1     do.  mess  pork, 


Richard  Lewis  Dr. 

To  2yds.  superfine  broadcloth  at  $6 
4  -  cassimere,  best  blk.  15s. 

1^  doz.  buttons,  5s.  6d. 

1     do.  small  do, 
4  skeins  silk, 
4  sticks  twist. 


By  cash  in  part, 


Cr. 


200 


cts. 


340 


462 


44 


90 


20 
15 
25 


10 
00 
00 


60 


10 


12 

10 

1 


00 
00 

37 
38 

25 


24 
15 


25 
00 


(2) 
Hallowell,  April  IS,  1827. 


231 


£= 

Thomas  Tilton. 
To  cash  in  part  of  the  balance  due 
his  order  paid  Samuel  Lane, 

A  nrjl   14  — 

Dr. 

him, 

$ 
50 
21 

cts. 
00 
50 

71 

50 

George  Simpson 
By  cash  (per  rect.) 

Anril  °4  ,  .            

Cr. 

41 

00 

Richard  Lewis                              Dr. 
To  5  pieces  India  cottons,            at  26s, 
7  yds.  cotton  cambrick,             4s.  6d. 
6  "  col'd  do.                               3s. 

Anril  rtci 

21 
5 
3 

67 
25 
00 

29 

92 

George  Simpson 
By  check  on  Gardiner  Bank  for  > 
one  hundred  and  fifty  dollars,     ) 

Anril  ^0 

Cr. 

150 

00 

John  Barton 
By  cash  rec'd  for  25bbls.  beef  sold 
at  $10,50cts. 

,  Tnnr    f\    L- 

Cr. 

} 

262 

50 

George  Simpson 
To  10  gallons  wine,                     at  $ 

Dr. 

1,20 

12 

00 

INDEX  TO  LEGER  NO  2. 


B  Barton  John, 

L  Lewis  Richard, 

P  Prince  Charles, 

R  Reed  William, 

S  Simpson  George, 

T  Tilton  Thomas, 


folio  I 

2 

2 

1 

1,2 

a 


232 


CO 
LEGER,  No  2. 


1827.    Dr.  George  Simpson, 


April  6 

To  balance, 

F 

ol. 

1 

$ 
200 

Cts. 
00 

Transferred  to  folio  2. 

200 

00 

1827. 

Dr.            John  Barton,     (1 

3; 

ith 

) 

April  6 

To  balance  for  wine, 

1 

I  $ 
346 

Cts. 
00 

1827. 

Dr.             William  Reed, 

1827. 


(l) 
LEGER,  No.  2. 

Contra 


233, 


Cr. 


April  14 
25 
June     1 

By  cash, 
By  check  on  Gardiner  Bank, 
By  balance  transferred  to  folio  2, 

Fol 
2 
2 

$ 
41 
150 
9 

Cts. 
00 
00 
00 

— 

—  —  , 

j 

200 

00 

1827. 

Contra                    Cr. 

April  30 

• 

By  cash, 

2 

262 

50 

1827. 

Contra                  Cr 

April  6  3 

3y  balance  due, 

1 

462 

44 

v  a 


234                                     (2) 
1827.        Dr.            Thomas  Tilton, 

April  13 

To  sundries, 

2 

$  i 
71 

as. 

50 

1827.        Dr.         Charles  Prince, 

AprilS 

To  sundries, 

1 

60 

10 

1827.         Dr.        Richard  Lewis,  (York.) 

April  8 
24 

To  sundries,        - 
To  sundries,        ... 

1 

2 

24 
29 

25 
92 

i 

1827.        Dr.             George  Simpson, 

June  1 
6 

To  balance  from  folio  1, 
To  wine,        .... 

2 

1     9 
12 

I 

00 
00 

(2)                                           235 
1827.             Contra                      Cr. 

April  6 

By  balance, 

1 

$ 
90 

as. 

00 

1827.             Contra                      Cr. 

1827.                 Contra                    Cr. 

April  8 

By  cash, 

1 

15 

00 

1827.                 Contra                    Cr. 

236  TABLES. 

A  TABLE  jor  reducing  Shillings  and  Pence  into 
and  Mills. 


SMI.      Shil.    Shil    Shil.    ShiL 

0 

1 

2 

3 

4 

5 

Pn.cts.m. 

cts. 

m. 

cts. 

m. 

cts.  m.  cts. 

m. 

cts. 

m. 

0 

16 

7 

.33 

3J50 

0|66 

7 

83 

3 

1 

1 

4 

18 

1|34 

7 

51 

468 

1 

84 

7 

2 

2 

8 

19 

536 

1 

52 

8|69 

5 

86 

1 

3 

4 

2 

20 

9|37 

5 

54 

2|70 

9|87 

5 

4 

5 

6 

22 

338 

9|55 

6|72 

388 

9 

5 

7 

0 

23 

7J40 

3 

57 

0|73 

7|90 

3 

6 

8 

3|25 

0|41 

6(58 

3|75 

0|91 

6 

7 

9 

7|26 

4 

43 

059 

7|76 

4 

93 

0 

8 

11 

1|27 

8 

44 

4|61 

1|77 

8 

94 

4 

9 

12 

o 

29 

0 

45 

8|62 

5|79 

295 

8 

10 

13 

9 

30 

6 

47 

2|63 

980 

6|97 

2 

11 

15 

332 

0|48 

6|65 

3|82 

0|98 

6 

EXAMPLE  - — Reduce  3s.  6d.  to  cents  and  mills.  Look  for  3s. 
at  the  head  of  the  column,  and  6  under  pence  at  the  left  hand 
side :  then  casting-  your  eye  along-  in  that  line  until  you  come  te 
the  Ssi  column,  you  have  58  cents  3  mills,  the  answer. 

Tables  of  the  value  of  the  Gold  Coins  of  Great  Britain^Franct  and 
Spain  according  to  the  act  of  Congress  of  April  29,  1816. 

[Gold  Coins  of  FRANCE.]  [Gold  Coins  of  SPAINJ 

grjct.  grs.ct.  pwt.  %  ct.     pwt.  $  cL  |  gr.  ct  gr.  ct.  pt.  g, ct.pt.  g  ct. 


f> 

3 

13 

47 

] 

0,87 

11 

9 

,60  I 

1 

3 

13 

45 

1  0.84 

11 

9,24 

2 

7 

14 

51 

2 

1,75 

12 

10 

>471 

2 

7 

14 

49 

2 

1,68 

12 

10,08 

3 

11 

15 

55 

3 

262 

13 

11 

,34 

3 

11 

15 

52 

3 

2,52 

13 

10,92 

4 

14 

16 

58 

4 

3,49 

14 

12 

,21  f 

4 

14 

16 

56 

4 

3,36 

14 

11,76 

5 

18 

17 

62 

5 

4,36 

15 

13,09  { 

5 

17 

17 

59 

5 

4,20 

15 

12,60 

6 

22 

18 

65 

6 

3,23 

16 

13 

:96{ 

6 

>I 

IS 

f>3 

6 

5,04 

16 

13,44 

7 

25 

19 

68 

7 

6,11 

17 

14,83  < 

7 

24119 

66 

7 

5,88 

17 

14,28 

8 

29 

20 

72 

8 

6,98 

18 

15,71  } 

8 

2820 

70 

8 

6,72 

18 

15,12 

9 

33 

21 

76 

9 

7,85 

19 

16,58  \ 

9 

32 

21 

73 

9 

7,56 

19 

15,96 

10 

3,6 

2280 

10|8,73 

20 

17 

,45  * 

10 

36 

22 

77 

108,402016,80 

11 

40 

23 

84 

i 

1  | 

39 

23 

SO 

12 

44 

1 

12 

42 

Gold  Coins  of  GREAT  BRITAIN  and  PORTUGAL. 

or 

ct. 

g-r 

ct  g 

r  ct. 

gr 

.ct. 

pwt 

dl.ct 

pwt 

dl.  ct 

spt 

dl.  cts 

pwt 

diets. 

& 
1 

3 

7 

25  1 

348 

19170 

I 

0,89 

1 

6,2^ 

\  \l 

»  11,55 

17 

15,11 

2 

7 

8 

29  1 

451 

20174 

2 

1,78 

6 

7,11 

\± 

t  12,44 

18 

16,00 

3 

11 

9 

33  | 

555 

21 

78 

3  |2,67      fl 

8,0( 

)ii 

>  13,33    19 

16,89 

4 

If 

10 

371 

659 

22 

81 

4 

3,55 

1C 

8,8< 

ne 

>  14,22 

20 

17,78. 

18 

11 

40  1 

763 

23 

85 

5 

4,44 

11 

9,78 

6 

22  12J44I18  67 

*6 

5,33 

12 

10,67 

FORMS  OF  NOTES  AND  BILLS  OF  EXCHANGE.   237 

FORMS  OF  NOTES,  BILLS,  RECEIPTS,   ifc. 

PROMISSORY  NOTE. 

Hallowell,  June  6,  1827. 

FOR  value  received,  I  promise  to  pay  one  hundred  and  twen- 
ty-one dollars  and  fifty  cents  to  George  Rich,  or  order,  in  sixty 
days,  with  interest.  HENRY  WEST. 

$121,50 

—  Witness,  Geo.  Spelman. 


I 


PROMISSORY  NOTE  BY  TWO  PERSONS. 

Hallowell,  June  6,  1827. 

For  value  received,  we  jointly  and  severally  promise  to  pay  fif- 
ty-six dollars  to  A.  B.  or  order,  on  demand,  with  interest. 

C.  Davis. 

$56,00  E.  Fox. 
Attest,  G,  Hill, 

NOTE  FOR  BORROWED  MONEY. 
Borrowed  and  received  of  C.  D.  forty-nine  dollars,  which  I 
promise  to  pay  on  demand.  E.  Fox. 

$49,00 

0^7"  A  promissory  note  having-  order  inserted,  may  be  endorsed 
from  one  person  to  another ;  and  if  value  received  is  not  mention- 
ed, it  is  of  no  force. 

INLAND  BILL  OF  EXCHANGE. 


$1000,00  Portland,  June  6, 1 827. 

Ten  days  after  sight,  pay  to  George  Brown  or  order,  one  thou- 
sand dollars,  for  value  received,  and  place  it  to  my  account  with- 
out further  advice,  (or  as  advised,)  from 

Your  humble  servant,  HENRT  WEST. 

To  Mr.  George  Rich,  Boston. 

FOREIGN  BILL  OF  EXCHANGE. 

EXCHANGE  for  £400  sterling. 

Hallowell,  June  7,  1827. 

Sixty  days  after  sight,  (or  at  usance,*)  pay  this  my  first  bill  of 
exchange,  second  and  third  of  the  same  tenor  and  date  not  paid, 
to  Mr.  George  Brown  or  his  order,  four  hundred  pounds  sterling 

*  Usance  is  a  customary  time  for  the  payment  of  foreign  bills  of 
exchange,circulating  from  one  nation  to  another;  and  varies  from 
80  to  90  days,  according  to  the  custom  of  different  countries. 


238  RECEIPTS BANK   DISCOUNT. 

(exchange  at  four  shilling's  and  sixpence  per  dollar)  for  value  re- 
ceived, and  place  it  (with  or  without  further  advice,)  to  the  ac- 
count of  Your  humble  servant. 

HENRY  WEST. 
Messrs.  Neil  fy  Thompson, 
Merchants,  Liverpool. 

RECEIPT  FOR  MONEY  PAID  ON  NOTE, 

Hallowell.Dec  6,  1827. — Received  from  William  Grant  (by  the 
hands  of  Thomas  Amory)  sixty-one  dollars  and  fifty  cents,  which 
is  endorsed  on  his  note  of  May  16th.  1825. 

SAMUEL  PRINCE. 
$61,50 


RECEIPT  FOR  MONEY  RECEIVED  ON  ACCOUNT. 

June  7,  1827 — Received  from  D.  E.  (by  the  hands  of  G    H.) 
forty  dollars  on  account.  L.  M. 

$40,00 


GENERAL  RECEIPT. 

June  7,  1827.— Received  of  N.  O.  ten  dollars  and  twenty-nine 
cents,  in  full  of  all  demands.  N.  B. 


$10,29 

N.  B — A  general  receipt  will  discharge  all  debts,  except  such 
as  are  on  specialty,  that  is,  bonds,  bills,  and  other  instruments 
that  may  properly  be  called  acts  or  deeds,  viz,  those  that  require 
to  be  executed  in  a  solemn  manner,  where  the  sealing  and  deliv- 
ery are  the  most  essential  parts  of  the  act,  and  on  that  account 
can  only  be  destroyed  or  cancelled  by  something  of  equal  force, 
viz,  some  other  specialty,  such  as  a  general  release,  &c.  Neither 
wUl  it  discharge  endorseable  promissory  notes,  or  Inland  bills. 

BANK  DISCOUNT. 

When  a  note  is  offered  at  a  bank  for  discount, two  endorsers  are 
generally  required,  to  the  first  of  whom  it  is  made  payable :  Thus 
A,  having  occasion  to  borrow  money,  procures  B  and  C  as  en- 
dorsers to  nig  note,  and  offers  it  for  discount  in  the  following  form. 


$500,00  Hallowell,  June  6, 1827. 

For  value  received,  I  promise  fro  pay  five  hundred  dollars  toB 
or  order,  at  the^Gardiner  Bank,  in  fifty -seven  days,  with  custom- 
ary grace. 

The  method  used  among  bankers  in  discounting  notes,  &c  is  to 


INVOICE   AND   ACCOUNT.  239 

find  the  interest  of  the  sum  from  the  date  of  the  note  to  the  time 
when  it  becomes  due,  including  the  days  of  grace  ;  the  interest 
thus  found  is  reckoned  the  discount,  and  is  taken  from  the  amount 
of  the  note  at  a  time,  before  the  person  receives  his  money. 

Grace  denotes  a  term  of  three  days,  which  custom  has  allowed 
to  the  borrower :  that  is,  though  the  note  becomes  due  in  fifty- 
seven  days,  he  may  withhold  payment  until  the  sixtieth,  for  which 
reason  the  interest  is  reckoned  for  sixty  days,  notwithstanding 
tne  note  should  be  paid  the  fifty-seventh  day. 


INVOICE  OF  GOODS. 

Boston,  June  6,  1827. 
Mr.  N.  BROWN  bought  of 

GEORGE  RICH. 

32  ells  Mode,                                              at  3s.  4d.  $17  78 

64  yds.  Striped  Nankins,         -        -             Is.  6d.  16  00 

28  *'  Calico, Is.  9d.  6  17 

4  pieces  Muslin,                            <•            30s.           -  20  00 

56  yds.  Cotton  Cassimere,        -                    2s.     -        -  18  67 

20  pieces  of  India  Cottons,          -        -        18s.         -  60  00 

25     "       plain  Nankins,                -               6s.  6d.         -  27  08 

2  doz.  cotton  Hose,  (Metis)                      66s.        -  22  00 

$189  70 
Rec'd  payment  by  his  Note  at  60  days, 

GEO.  HIGH. 


ACCOUNT  RENDERED. 

MR.  RICHARD  LEWIS, 
1827.  To  A- 

April  1,     To  2yds.  Superfine  Cloth, 

4    "    blk.  Cassimere, 

1J  doz.  Buttons, 

1     «     small  do. 

4  skeins  silk, 

4  sticks  of  twist, 
24,          5  pieces  Ind.  Cottons,  22yds.  ea. 

7  yds.  Cotton  Cambric, 

6  "     col'd  do. 
May  18,          14    «    do.  do. 

7  "    Linen, 

1  oz.  Thread,  No.  40,        -        * 

2  pair  Morocco  Shoes,        -  1,08 
1  oz.  Indigo, 

Hallowell,  Dec.  6, 1827.  $70  43 

Rec'd  payment  for  A —  B . 

GEORGE  NORTH. 


240 


ACCOUNT  CURRENT. 


o 


' 


**£; 


ELISHA  NAYSON.5 

•*$  O 


M305D91 

QA  102 


THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


Y v4  'o  V 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 


AN  INITIAL  PINE  OF  25  CENTS 

WILL  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  SO  CENTS  ON  THE  FOURTH 
DAY  AND  TO  $1.OO  ON  THE  SEVENTH  DAY 
OVERDUE. 


/  J 

7  r 

^"^ 

7l 


/<// 


B  /73?5 


cn 
0 


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